4 pounds of oranges costs $ 12 . What is the unit price per pound?

Answer:

The solution is:

(1) 0.0104

(2) 0.0008

Step-by-step explanation:

Given:

Mean,

[tex]\mu = 43.7[/tex]

Standard deviation,

[tex]\sigma = 1.6[/tex]

(1)

⇒ [tex]P(X<40) = P(\frac{x-\mu}{\sigma}<\frac{40-43.7}{1.6} )[/tex]

                      [tex]=P(z< - 2.3125)[/tex]

                      [tex]=P(z<-2.31)[/tex]

                      [tex]=0.0104[/tex]

(2)

As we know,

n = 15

⇒ [tex]P(\bar X > 45)= P(\frac{\bar x - \mu}{\frac{\sigma}{\sqrt{n} } } >\frac{45-43.7}{\frac{1.6}{\sqrt{15} } } )[/tex]

                      [tex]=P(z> 3.15)[/tex]

                      [tex]=1-P(z<3.15)[/tex]

                      [tex]=1-0.9992[/tex]

                      [tex]=0.0008[/tex]

Answer:

a) 0.18 = 18% probability a randomly chosen business customer flying with Global is on a jumbo jet.

b) 0.3 = 30% probability a randomly chosen non-business customer flying with Global is on an ordinary jet.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

P(B|A) is the probability of event B happening, given that A happened.

[tex]P(A \cap B)[/tex] is the probability of both A and B happening.

P(A) is the probability of A happening.

Question a:

Event A: Jumbo

Event B: Business

60% of its capacity is provided on jumbo jets.

This means that [tex]P(A) = 0.6[/tex]

On jumbo jets, 30% of the passengers are on business

This means that [tex]P(B|A) = 0.3[/tex]

Desired probability:

We want to find [tex]P(A \cap B)[/tex], so:

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

[tex]P(A \cap B) = P(B|A)P(A) = 0.6*0.3 = 0.18[/tex]

0.18 = 18% probability a randomly chosen business customer flying with Global is on a jumbo jet.

b) What is the probability a randomly chosen non-business customer flying with Global is on an ordinary jet?

Event A: Ordinary

Event B: Non-business

60% of its capacity is provided on jumbo jets.

So 100 - 60 = 40% are ordinary, which means that [tex]P(A) = 0.4[/tex]

On ordinary jets 25% of the passengers are on business.

So 100 - 25 = 75% are non-business, that is [tex]P(B|A) = 0.75[/tex]

Desired probability:

We want to find [tex]P(A \cap B)[/tex], so:

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

[tex]P(A \cap B) = P(B|A)P(A) = 0.75*0.4 = 0.3[/tex]

0.3 = 30% probability a randomly chosen non-business customer flying with Global is on an ordinary jet.

can you zoom in on my pic more or no does it say 1/z

Answer:

Option A. Reciprocal

Answered by GAUTHMATH

Answer:

The answer is "18".

Step-by-step explanation:

In the given graph by concluding we observe that  on the x-axis, one step is 2 units, and when we half each of the steps it will= 1 unit

[tex]\therefore\\\\CD = distance\ from\ -(8+1)\ to\ (8+1)\ = \text{distance between} -9 \ to\ 9\ = 18[/tex]

Answer:

The answer is A

Step-by-step explanation:

(-2,-3) (3,-4) (0,-1) (-7,3)

inverse means the opposite signs

(2,3) (-3,4) (0,1) (7,-3)

Answer:

6+7=13

13+8=21

32+21=52

11+34=46

31+03=34

Step-by-step explanation:

im not sure in the 31+03

Answer:

4 scoops of ice cream

Step-by-step explanation:

Plug in the total cost, number of scoops of nuts, and number of liquid toppings into the formula. Then, solve for s:

C = 0.50s + 0.35n + 0.25t

3.55 = 0.50s + 0.35(3) + 0.25(2)

3.55 = 0.50s + 1.05 + 0.5

3.55 = 0.50s + 1.55

2 = 0.50s

4 = s

So, the sundae had 4 scoops of ice cream.

9514 1404 393

Answer:

  (9√3 -3π/2) ft^3 ≈ 10.88 ft^3

Step-by-step explanation:

The area of the hexagon is given by the formula ...

  A = (3/2)√3·s^2 . . . . for side length s

The area of the hexagonal face of this solid is ...

  A = (3/2)√3·(2 ft)^2 = 6√3 ft^2

__

The area of the circular hole in the hexagonal face is ...

  A = πr^2

The radius is half the diameter, so is r = (2 ft)/2 = 1 ft.

  A = π(1 ft)^2 = π ft^2

Then the area of the "solid" part of the face of the figure is ...

  A = (6√3 -π) ft^2

__

The volume is ...

  V = Bh . . . . . where B is the area of the base of the prism, and h is its height

  V = ((6√3 -π) ft^2)(3/2 ft) = (9√3 -3π/2) ft^3 ≈ 10.88 ft^3

Step-by-step explanation:

1 since a tank holds 121/2 gallons of gas 121/2 multiplied by two it gives 25 -30 it gives 5 450 divided by two it gives 225-25 i guess.

Answer:

A.

Step-by-step explanation:

A. is the answer because they list all of the sandwiches, soups, and beverages with every possible combination.

Answer: 6.9

Step-by-step explanation:

Answer:

-2.9 > w

Step-by-step explanation:

1.8>4.7+w

Subtract 4.7 from each side

1.8-4.7>4.7-4.7+w

-2.9 > w

Answer:

w = -2.9

Step-by-step explanation:

Answer:

ACB = JCB

Step-by-step explanation:

ASA means angle - (included) side - angle.

we have one angle confirmed.

we have the connected side BC confirmed (the diagram shows that the side is shared, so it is not only congruent, it is actually identical).

now we need confirmation for the angle at the other end point of that side.

Answer: [tex]10,000\ lb.ft[/tex]

Step-by-step explanation:

Given

Initial weight of the bucket is [tex]145\ lb[/tex]

It is lifted at constant rate and rate of sand escaping is [tex]0.5\ lb/ft[/tex]

At any height weight of the sand is [tex]w(h)=145-0.5h[/tex]

Work done is given by the product of applied force and displacement or the area under weight-displacement graph

from the figure area is given by

[tex]\Rightarrow W=\int_{0}^{80}\left ( 145-0.5h \right )dh\\\\\Rightarrow W=\left | 145h-\dfrac{0.5h^2}{2} \right |_0^{80}\\\\\Rightarrow W=\left [ 145\times 80-\dfrac{0.5(80))^2}{2} \right ]-0\\\\\Rightarrow W=11,600-1600\\\\\Rightarrow W=10,000\ lb.ft[/tex]

Answer:

8 cm^2

Step-by-step explanation:

If the triangle is isosceles the sides are the same

Let the sides be x

We know that we can use the Pythagorean theorem

a^2+b^2 = c^2 where a and b are the sides and c is the hypotenuse

x^2+x^2 = (4 sqrt(2))^2

2x^2 =16(2)

2x^2 = 32

Divide by 2

2x^2/ 32/2

x^2 = 16

Taking the square root of each side

sqrt(x^2) = sqrt(16)

x = 4

The area of the triangle is

A =1/2 bh

A = 1/2 (4) (4)

A = 1/2(16)

A = 8

Answer:

8

Step-by-step explanation:

a^2 + b^2 = c^2

c = [tex]4\sqrt{2}[/tex]

[tex]c^{2} = 32[/tex]

a^2 + b^2 = 32

a=b (isosceles triangle)

a=b=4

base = 4

height = 4

area = 1/2 bh = 1/2(4)(4) = 8

Answer:

probability of the product of the chosen integers being a multiple of 3 is P(E)= 1 - (91/285)

=194/285 or 0.6807.

Step-by-step explanation:

The sample space of all possible choices of three integers from the set {1,2,…..,20} has C(20,3) = (20×19×18)/3! elements.

The complement of the event space E consists of all possible choices of three integers from the complement of the set of all the multiples of 3 in the above set because the product of the chosen integers is a multiple of 3 if and only if at least one of them is a multiple of 3. Hence we have to choose three elements from the set {1,2,4,5,7,8,10,11,13,14,16,17,19,20} which has 14 elements.

Hence

|E'| = C(14,3)

= 14×13×12/3!.

Therefore probability P(E')

= |E'|/|S|

= (14×13×12)/(20×19×18)

= (14×13×2)/(20×19×3)

=(7×13)/(5×19×3)

= 91/285.

Therefore the required probability of the product of the chosen integers being a multiple of 3 is P(E)= 1 - (91/285)=194/285 or 0.6807.

Answer:

-49

8

Step-by-step explanation:

-196/4 =

-49

(3+u)^2

-------------

8

Let u=5

(3+5)^2

-------------

8

(8)^2

-------------

8

64

-----

8

8

Answer:

15

Step-by-step explanation:

1-0 =1

3-1 =2

6-3=3

10-6=4

We are adding 1 more each time

10+5 = 15

Solution :

We have given two baskets :

[tex]$H_1$[/tex] : 8 apples + 2 bananas + 6 cantaloupes = 16 fruits

[tex]$H_2$[/tex] : 6 apples + 4 bananas = 10 fruits

A fair coin is made to flipped. If the [tex]\text{flip}[/tex] results is head, then the fruit is selected from a basket [tex]$H_1$[/tex].

If the flip results in tail, then the fruit is selected from the basket [tex]$H_2$[/tex].

Probability of head P(H) = [tex]1/2[/tex]

Probability of tail P(T) = [tex]1/2[/tex]

if given event is :

B = selected fruit is BANANA

We have to calculate : P(T|B)

Probability of banana if the flip results is head P(B|H) = [tex]$\frac{2}{16}$[/tex]

Probability of banana if the flip results is tail P(B|T) = [tex]$\frac{4}{10}$[/tex]

From the Bayes' theorem :

Probability of flip results is tail when selected fruit is BANANA.

[tex]$P(T|B) = \frac{P(B|T)\ P(T)}{P(B|T) \ P(T) + P(B|H)\ P(H)}$[/tex]

            [tex]$=\frac{\frac{4}{10} \times \frac{1}{2}} {\frac{4}{10} \times \frac{1}{2} + \frac{2}{16} \times \frac{1}{2}}$[/tex]

            [tex]$=\frac{\frac{1}{5}}{\frac{1}{5}+\frac{1}{16}}$[/tex]

            [tex]$=\frac{\frac{1}{5}}{\frac{21}{80}}$[/tex]

            [tex]$=\frac{16}{21}$[/tex]

∴  [tex]$P(\ T|B\ )=\frac{16}{21}$[/tex]

Answer:

17690

Step-by-step explanation:

The amount Michael owes is given by :

y = - 290x +18,560

x = number of months ; y = amount owed

Amount owed after 3 months :

Put x = 3 in the equation :

y= - 290(3) +18,560

y = - 870 + 18560

y = 17690

According to the general form of a linear equation :

y = mx + c

Where, m = slope ; c = intercept

The slope = - 290 ; Intercept = 18560

Slope is the change in y per unit change in x ; The slope means that amount owed, y decreases by 290 per month

The intercept shows that the original amount borrowed is 18560

Amount Michael owes his father decreases by 290 per month.

Hello there!

Previously, we learnt that to solve the equation, we have to isolate the sin, cos, tan, etc first.

First Question

The first question has sin both sides. Notice that if we move sin(theta) to left. We get:-

[tex] \displaystyle \large{2 {sin}^{2} \theta - sin \theta = 0}[/tex]

We can common factor out the expression.

[tex] \displaystyle \large{sin \theta(2sin \theta - 1) = 0}[/tex]

It is a trigonometric equation in quadraric pattern.

We consider both equations:-

First Equation

[tex] \displaystyle \large{sin \theta = 0}[/tex]

Remind that sin = y. When sin theta = 0. It means that it lies on the positive x-axis.

We know that 0 satisfies the equation, because sin(0) is 0.

Same goes for π as well, but 2π does not count because the interval is from 0 ≤ theta < 2π.

Hence:-

[tex] \displaystyle \large { \theta = 0,\pi}[/tex]

Second Equation

[tex] \displaystyle \large{2sin \theta - 1 = 0}[/tex]

First, as we learnt. We isolate sin.

[tex] \displaystyle \large{sin \theta = \frac{1}{2} }[/tex]

We know that, sin is positive in Quadrant 1 and 2.

As we learnt from previous question, we use π - (ref. angle) to find Q2 angle.

We know that sin(π/6) is 1/2. Hence π/6 is our reference angle. Since π/6 is in Q1, we only have to find Q2.

Find Quadrant 2

[tex] \displaystyle \large{\pi - \frac{\pi}{6} = \frac{6\pi}{6} - \frac{\pi}{6} } \\ \displaystyle \large{ \frac{5\pi}{6} }[/tex]

Hence:-

[tex] \displaystyle \large{ \theta = \frac{\pi}{6} , \frac{5\pi}{6} }[/tex]

Since both first and second equations are apart of same equation. Therefore, mix both theta from first and second.

Therefore, the solutions to the first question:-

[tex] \displaystyle \large \boxed{ \theta = 0,\pi, \frac{\pi}{6} , \frac{5\pi}{6} }[/tex]

Second Question

This one is a reciprocal of tan, also known as cot.

[tex] \displaystyle \large{cot3 \theta = 1}[/tex]

For this, I will turn cot to 1/tan.

[tex] \displaystyle \large{ \frac{1}{tan3 \theta} = 1}[/tex]

Multiply whole equation by tan3 theta, to get rid of the denominator.

[tex] \displaystyle \large{ \frac{1}{tan3 \theta} \times tan3 \theta = 1 \times tan3 \theta } \\ \displaystyle \large{ 1= tan3 \theta }[/tex]

We also learnt about how to deal with number beside theta.

We increase the interval, by multiplying with the number.

Since our interval is:-

[tex] \displaystyle \large{0 \leqslant \theta < 2\pi}[/tex]

Multiply the whole interval by 3.

[tex] \displaystyle \large{0 \times 3 \leqslant \theta \times 3 < 2\pi \times 3} \\ \displaystyle \large{0 \leqslant 3 \theta < 6\pi }[/tex]

We also know that tan is positive in Quadrant 1 and Quadrant 3.

and tan(π/4) is 1. Therefore, π/4 is our reference angle and our first theta value.

When we want to find Quadrant 3, we use π + (ref. angle).

Find Q3

[tex] \displaystyle \large{\pi + \frac{\pi}{4} } = \frac{5\pi}{4} [/tex]

Hence, our theta values are π/4 and 5π/4. But that is for [0,2π) interval. We want to find theta values over [0,6π) interval.

As we learnt previously, that we use theta + 2πk to find values that are in interval greater than 2π.

As for tangent, we use:-

[tex] \displaystyle \large{ \theta + \pi k = \theta}[/tex]

Because tan is basically a slope or line proportional graph. So it gives the same value every π period.

Now imagine a unit circle, and make sure to have some basic geometry knowledge. Know that when values addition by 180° or π would give a straight angle.

We aren't using k = 1 for this because we've already found Q3 angle.

Since we know Q1 and Q3 angle in [0,2π).

We can also use theta + 2πk if you want.

First Value or π/4

[tex] \displaystyle \large{ \frac{\pi}{4} + 2\pi = \frac{9\pi}{4} } \\ \displaystyle \large{ \frac{\pi}{4} + 4\pi = \frac{17\pi}{4} }[/tex]

Second Value or 5π/4

[tex] \displaystyle \large{ \frac{5\pi}{4} + 2\pi = \frac{13\pi}{4} } \\ \displaystyle \large{ \frac{5\pi}{4} + 4\pi = \frac{21\pi}{4} }[/tex]

Yes, I use theta + 2πk for finding other values.

Therefore:-

[tex] \displaystyle \large{3 \theta = \frac{\pi}{4} , \frac{5\pi}{4} , \frac{9\pi}{4}, \frac{17\pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} }[/tex]

Then we divide every values by 3.

[tex] \displaystyle \large \boxed{\theta = \frac{\pi}{12} , \frac{5\pi}{12} , \frac{9\pi}{12}, \frac{17\pi}{12} , \frac{13\pi}{12} , \frac{21\pi}{12} }[/tex]

Let me know if you have any questions!

Answer:

can you please send the picture of the diagram.

Step-by-step explanation:

Answer:

ABC:

AB-5 units

BC-12.65 units

AC-15 units

FGH:

FG-5 units

GH-12.65 units

FH-15 units

Step-by-step explanation:

PLATO SAMPLE ANSWER

Answer:

The cube root parent function:

Horizontally stretched by a factor of 4:

Translated 5 units right:

Translated 3 units up:

Step-by-step explanation:

a) Weight of the water:

Volume of bottle:

b) Weight of brine:

Density of brine:

Answer:

40141

Step-by-step explanation:

Answer:

The x-intercept is -8, the y-intercept is 4, and the slope is 1/2.

Step-by-step explanation:

The x-intercept is -8, the y-intercept is 4, and the slope is 1/2.

Answer: See below

Concept:

There are different forms of linear equations:

Solve:

Given: y - 3 = 1/2 (x + 2)

Here, we can see the linear equation is in the form of point-slope form.

x₁ = -2

y₁ = 3

m = 1/2

Point included in the graph = (-2, 3)

Slope of the graph = 1/2

Hope this helps!! :)

Please let me know if you have any questions

Step-by-step explanation:

here is your answer it may help you

There are 700 don't like to admit in both faculties.

The process of combining two or more numbers is called the Addition. The 4 main properties of addition are commutative, associative, distributive, and additive identity.

Given that;

In a survey, survey of 1200 students who have passed SEE 150 like to admit in science Faculty 600 in humanity first and 240 like to admit either of faculties and the rest were found not to be admitted in both faculties​

Hence, not to be admit in any faculty =   N

Science +  Humanity -  Either   + None  = 200

=> 150 + 600 - 240 + N = 1200

=> 750 - 240 + N = 1200

=> N = 700

Hence, 700 don't like to admit in both faculties.

Learn more about the addition visit:

https://brainly.com/question/25421984

#SPJ2

Answer:

By the Empirical Rule, approximately 68% of the bulbs have lifetimes that lie within 1 standard deviation to either side of the mean.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

What percentage of the bulbs have lifetimes that lie within 1 standard deviation to either side of the mean?

By the Empirical Rule, approximately 68% of the bulbs have lifetimes that lie within 1 standard deviation to either side of the mean.