Please explain the answer

Answer:

Test statistic = 0.63

Pvalue = 0.555

A. Fail to reject H0. The data does not suggest a significant mean difference in the average number of accidents after information was added to road signs.

Step-by-step explanation:

Given :

Before 13 22 65 123 56 63

After_ 14 21 43 84 75 72

To perform a paired t test :

H0 : μd = 0

H1 : μd ≠ 0

We obtain the difference between the two dependent sample readings ;

Difference, d = -1, 1, 22, 39, -19, -9

The mean of difference, Xd = Σd/ n = 33/6 = 5.5

The standard deviation, Sd = 21.296 (calculator).

The test statistic :

T = Xd ÷ (Sd/√n) ; where n = 6

T = 5.5 ÷ (21.296/√6)

T = 5.5 ÷ 8.6940555

T = 0.6326

The Pvalue : Using a Pvalue calculator ;

df = n - 1 = 6 - 1 = 5

Pvalue(0.6326, 5) = 0.5548

Decision region :

Reject H0 ; If Pvalue < α; α = 0.05

Since 0.5548 > 0.05 ; we fail to reject the Null and conclude that the data does not suggest a significant mean difference in the average number of accidents after information was added to road signs.

Answer:

-1.683

Step-by-step explanation:

Given :

Group 1 :

x1 = 667 ; n1 = 10 ; s1 = 20

Group 2 :

x2 = 679 ; n2 = 14 ; s2 = 15

The test statistic assuming equal variance :

x1 - x2 / √[Sp² * (1/n1 + 1/n2)]

sp² = [(n1 - 1)*s1² + (n2 - 1)*s2²] ÷ (n1 + n2 - 2)

Sp² = [(10 - 1)*20² + (14 - 1)*15²] = 296.59

Test statistic =

(667 - 679)/ √[296.59 * (1/10 + 1/14)]

-12 / 7.1304978

Test statistic = - 1.682

Answer:

The answer is [tex]x=-16[/tex] ±[tex]2\sqrt{3}[/tex] in exact form or [tex]x=-12.535898[/tex], [tex]x=-19.464102[/tex] in decimal form.

Step-by-step explanation:

To solve this problem, start by moving all terms to the left side of the equation and simplify. Simplify the equation by subtracting 12 from both sides of the equation and squaring [tex]x+16[/tex], which will look like [tex]x^{2} +32x+256-12=0[/tex]. Next, simplify the equation again, which will look like [tex]x^{2} +32x+244=0[/tex].

Then, use the quadratic formula to find the solutions. The quadratic formula looks like[tex]\frac{-b(+-)\sqrt{b^{2}-4ac } }{2a}[/tex].

For this problem, the quadratic variables are as follows:

[tex]a=1[/tex]

[tex]b=32[/tex]

[tex]c=244[/tex]

The next step is to substitute the values [tex]a=1[/tex], [tex]b=32[/tex], and [tex]c=244[/tex] into the quadratic formula and solve. The quadratic formula will look like [tex]\frac{-32(+-)\sqrt{32^2-4(1*244)} }{2*1}[/tex]. To simplify the equation, start by simplifying the numerator, which will look like [tex]x=\frac{-32(+-)4\sqrt{3} }{2*1}[/tex]. Then, multiply 2 by 1 and simplify the equation, which will look like [tex]x=-16(+-)2\sqrt{3}[/tex]. The final answer is [tex]x=-16[/tex] ±[tex]2\sqrt{3}[/tex] in exact form. In decimal form, the final answer is [tex]x=-12.535898[/tex], [tex]x=-19.464102[/tex].  

Answer:

The p-value of the test is 0.1469 > 0.05, which means that there is no reason to believe that the proportion of adults favoring capital punishment today has increased, using a 0.05 level of significance.

Step-by-step explanation:

Suppose that, in the past, 40% of all adults favored capital punishment. Test if the proportion has increased:

At the null hypothesis, we test if the proportion is still of 40%, that is:

[tex]H_0: p = 0.4[/tex]

At the alternative hypothesis, we test if the proportion has increased, that is, is greater than 40%, so:

[tex]H_1: p > 0.4[/tex]

The test statistic is:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.

0.4 is tested at the null hypothesis:

This means that [tex]\mu = 0.4, \sigma = \sqrt{0.4*0.6}[/tex]

Random sample of 15 adults, 8 favor capital punishment.

This means that [tex]n = 15, X = \frac{8}{15} = 0.5333[/tex]

Value of the test statistic:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{0.5333 - 0.4}{\frac{\sqrt{0.4*0.6}}{\sqrt{15}}}[/tex]

[tex]z = 1.05[/tex]

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion of 0.5333 or more, which is 1 subtracted by the p-value of z = 1.05.

Looking at the z-table, z = 1.05 has a p-value of 0.8531.

1 - 0.8531 = 0.1469.

The p-value of the test is 0.1469 > 0.05, which means that there is no reason to believe that the proportion of adults favoring capital punishment today has increased, using a 0.05 level of significance.

Answer:

115.625

PRT/100

There's nothing preventing us from computing one integral at a time:

[tex]\displaystyle \int_0^{2-x} xyz \,\mathrm dz = \frac12xyz^2\bigg|_{z=0}^{z=2-x} \\\\ = \frac12xy(2-x)^2[/tex]

[tex]\displaystyle \int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy = \frac12\int_0^{1-x}xy(2-x)^2\,\mathrm dy \\\\ = \frac14xy^2(2-x)^2\bigg|_{y=0}^{y=1-x} \\\\= \frac14x(1-x)^2(2-x)^2[/tex]

[tex]\displaystyle\int_0^1\int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx[/tex]

Expand the integrand completely:

[tex]x(1-x)^2(2-x)^2 = x^5-6x^4+13x^3-12x^2+4x[/tex]

Then

[tex]\displaystyle\frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx = \left(\frac16x^6-\frac65x^5+\frac{13}4x^4-4x^3+2x^2\right)\bigg|_{x=0}^{x=1} \\\\ = \boxed{\frac{13}{240}}[/tex]

Answer:

x = 20

y = 10

it's a 30-60-90 triangle

k = -3

Step-by-step explanation:

Let's rewrite the equations in their slope-intercept forms:

[tex]y = \frac{1}{3}x + 2\:\:\:\:\:\:\:(1)[/tex]

[tex]y = kx + 12\:\:\:\:\:(2)[/tex]

In order for Eqn(2) to be perpendicular to Eqn(1), the slope of Eqn(2) must be the negative reciprocal of that of Eqn(1). Therefore, [tex]k = -3[/tex].

Answer:

[tex]\displaystyle \frac{1}{9}[/tex]

Step-by-step explanation:

Hi there!

This question is asking us what the value of y is when x is -2, hence the point (-2,y).

[tex]y=3^x[/tex]

To find y, replace x in the equation with -2 and evaluate:

[tex]y=3^-^2[/tex]

When [tex]a^-^n[/tex] where n>0, [tex]a^-^n=\displaystyle \frac{1}{a^n}[/tex]:

[tex]y=\displaystyle \frac{1}{3^2} \\\\y=\displaystyle \frac{1}{9}[/tex]

I hope this helps!

Answer:

b. 300x1 + 125x2 < = 4,000

Step-by-step explanation:

Maximum of 4,000 hours

This means that the the total amount of labor has to be of at most 4,000, that is:

[tex]T \leq 4000[/tex]

300 ping-pong balls (x1) or 125 wiffle balls (x2) can be produced per hour of labor

The total amount of labor is:

[tex]T = 300x1 + 125x2[/tex]

Uniting the two equations:

[tex]T \leq 4000[/tex]

[tex]300x1 + 125x2 \leq 4000[/tex]

And thus the correct answer is given by option b.

Answer:

Step-by-step explanation:

16x-8y = 16 ⇒ 8x - 4y = 8, which is identical to the second equation.

The equations are equivalent, so there are an infinite number of solutions.

Answer:

Hello,

Step-by-step explanation:

Let say t the time the hose is left running

235 ≤ t < 245 (in min)

Let say d the amount of water coming out of the hose each minute

2.05 ≤ d < 2.15 (why d : débit in french)

235*2.05 ≤ t*d < 245*2.15

481.75 ≤ t*d < 526.75    (litres)

Answer:

Lower bound: [tex]495\; \rm L[/tex] (inclusive.)

Upper bound: [tex]505\; \rm L[/tex] (exclusive.)

Step-by-step explanation:

The amount of water from the hose is the product of time and the rate at which water comes out.

When multiplying two numbers, the product would have as many significant figures as the less accurate factor.

In this example, both factors are accurate to two significant figures. Hence, the product would also be accurate to two significant figures. That is:

[tex]240 \times 2.1 = 5.0 \times 10^{2}\; \rm L[/tex] ([tex]500\; \rm L[/tex] with only two significant figures.)

Let [tex]x[/tex] denote the amount of water in liters. For [tex]x\![/tex] to round to [tex]5.0 \times 10^{2}\; \rm L[/tex] only two significant figures are kept, [tex]495 \le x < 505[/tex]. That gives a bound on the quantity of water from the hose.

Answer:

b

Step-by-step explanation:

the graph gets translated 5 units above its parent graph of y = x

Answer:

Yes. the Psychologist can confirm his theory that the attitudes have changed over time, based on the first and second surveys.

The expected value for the number of respondents who are optimistic is:

= 16.25

Step-by-step explanation:

Attitude                   1st Survey      2nd Survey

Optimistic                     7%                    6%

Slightly Optimistic       9%                     6%

Slightly Pessimistic    31%                   37%

Pessimistic                53%                   51%

Expected value of optimistic respondents:

Attitude                    

Optimistic      Expected Value

1st Survey        8.75 (250 * 7% * 50%)  

2nd Survey      7.50 (250 * 6% * 50%)

Total EV         16.25

Answer:

top right

Step-by-step explanation:

roots of an equation = x-intercepts

Answer:

top right is the answer from my calculatins

Answer:  [tex]M = \pm\sqrt{\frac{3K}{4(5+7NK)}}\\\\[/tex]

======================================

Work Shown:

[tex]\frac{5}{K} = \frac{3}{4M^2} - 7N\\\\\frac{5}{K}+7N = \frac{3}{4M^2}\\\\\frac{5}{K}+7N*\frac{K}{K} = \frac{3}{4M^2}\\\\\frac{5}{K}+\frac{7NK}{K} = \frac{3}{4M^2}\\\\\frac{5+7NK}{K} = \frac{3}{4M^2}\\\\4M^2*(5+7NK) = K*3\\\\M^2 = \frac{3K}{4(5+7NK)}\\\\M = \pm\sqrt{\frac{3K}{4(5+7NK)}}\\\\[/tex]

Other forms are possible, but those forms would be equivalent to what is shown above.

The probabilities we found in this exercise are.

In this exercise, probability concepts are used.

A probability is the number of desired outcomes divided by the number of total outcomes.

Total number of countries:

23 + 12 + 47 + 44 + 54 + 14 = 194

Let A = the event that a country is in Asia.

44 of the 194 countries are in Asia, thus:

[tex]P(A) = \frac{44}{194} = 0.2268[/tex]

0.2268 = 22.68% probability that a country is in Asia.

Let E = the event that a country is in Europe.

47 out of 194 countries are in Europe, thus:

[tex]P(E) = \frac{47}{194} = 0.2423[/tex]

0.2423 = 24.23% probability that a country is in Europe.

Let F = the event that a country is in Africa.

54 out of 194 countries are in Africa, thus:

[tex]P(F) = \frac{54}{194} = 0.2784[/tex]

0.2784 = 27.84% probability that a country is in Africa.

Let N = the event that a country is in North America.

23 out of 194 countries are in North America, thus:

[tex]P(N) = \frac{23}{194} = 0.1186[/tex]

0.1186 = 11.86% probability that a country is in North America.

Let O = the event that a country is in Oceania.

14 out of 194 countries are in Oceania, thus:

[tex]P(O) = \frac{14}{194} = 0.0722[/tex]

0.0722 = 7.22% probability that a country is in Oceania.

Let S = the event that a country is in South America.

12 out of 194 countries are in South America, thus:

[tex]P(S) = \frac{12}{194} = 0.0619[/tex]

0.0619 = 6.19% probability that a country is in South America.

18. What is the probability of drawing a red card in a standard deck of 52 cards?

In a standard deck of 52 cards, 26 are red, and thus:

[tex]p = \frac{26}{52} = 0.5[/tex]

0.5 = 50% probability of drawing a red card in a standard deck of 52 cards.

19. What is the probability of drawing a club in a standard deck of 52 cards?

In a standard deck of 52 cards, 13 are clubs, and thus:

[tex]p = \frac{13}{52} = 0.25[/tex]

0.25 = 25% probability of drawing a club in a standard deck of 52 cards.

For more about probabilities, you can check https://brainly.com/question/24104122

Answer:     The answer is -2

Step-by-step explanation:

that is where the line starts

hope this helped

Answer:

x > -2

Step-by-step explanation:

The domain is the values that the input can take

x goes from -2( not including -2) to infinity

x > -2

Answer:

mean: 84

median: 85

mode: 85

explainations for d and e below.

Step-by-step explanation:

mean:

(85+80+96+72+78+85+92)/7 = 544/7 = 84

median:

sort them in order from least to greatest first

72, 78, 80, 85, 85, 92, 96

find the middle number in that order, that's your median. in this case, it's 85.

mode:

number that appears the most in the order; the frequency. in this case as well, it's also 85.

answer for d:

the reason why the mean is lower than the median is because of the fact we were figuring out the average and had to divide, unlike the median where it's just the middle of the order we have.

answer for e:

(85+80+96+72+78+85+92+100)/8 = 644/8 = 80.5%

p.s: please subscribe to #gauthmath# sub reddit if you can for more help.

Answer:

b or a

Step-by-step explanation:

Answer:

Solution given:

equation is:

x²-40x+A=0

when A=200

equation becomes

x²-40x+200=0

Comparing above equation with ax²+bx+c=0 we get

a=1

b=-40

c=200

x=[tex]\displaystyle \frac{-b±\sqrt{b²-4ac}}{2a}[/tex]

substituting value

x=[tex]\displaystyle \frac{-*-40±\sqrt{(-40)²-4*1*200}}{2*1}[/tex]

x=[tex]\displaystyle \frac{40±\sqrt{800}}{2}[/tex]

x=[tex]\displaystyle \frac{40±20\sqrt{2}}{2}[/tex]

taking positive

x=[tex]\displaystyle \frac{40+20\sqrt{2}}{2}[/tex]

x=34.14

taking negative

x=[tex]\displaystyle \frac{40-20\sqrt{2}}{2}[/tex]

x=5.86

Given:

The figures of two polygons.

To find:

Whether the polygons are similar and then find the scale factor (if similar).

Solution:

From the given figures it is clear that both polygons are rectangles and their all interior angles are right angles.

The ratio of their longer sides:

[tex]\dfrac{32}{26}=\dfrac{16}{13}[/tex]

The ratio of their shorter sides:

[tex]\dfrac{18}{12}=\dfrac{3}{2}[/tex]

Since the ratio of their corresponding sides are not equal, therefore the two polygons are not similar.

Therefore the required solutions are:

Similar : No

Similarity statement : None

Scale factor : None

The focus here is the use of "Compounding interest rate" and these entails addition of interest to the principal sum of the deposit.

The below calculation is to derive maturity value when annual rate of 3.1% is applied.

Principal = $5,400

Annual rate = 3.1% semi-annually for 1 years

A = P(1+r/m)^n*t where n=1, t=2

A = 5,400*(1 + 0.031/2)^1*2

A = 5,400*(1.0155)^2

A = 5,400*1.03124025

A = 5568.69735

A = $5,568.70.

In conclusion, the accrued value she will get after one years for this account is $5,568.70,

- The below calculation is to derive maturity value when the amount compounds continuously at an annual rate of 4%

Principal = $5,400

Annual rate = 4% continuously

A = P.e^rt where n=1

A = 5,400 * e^(0.04*1)

A = 5,400 * 1.04081077419

A = 5620.378180626

A = 5620.378180626

A = $5,620.39.

In conclusion, the accrued value she will get after one years for this account is $5,620.39.

Referring to how much would Tyra's balance be from Account 2 over 3.7 years. It is calculated as follows:

Annual rate = 4% continuously

A = P.e^rt where n=3.7

A = 5,400 * e^(0.04*3.7)

A = 5,400 * e^0.148

A = 5,400 * 1.15951289636

A = 6261.369640344

A = $6,261.37

Therefore, the accrued value she will get after 3.7 years for this account is $6,261.37

Learn more about Annual rate here

brainly.com/question/14170671

Answer:

$170 Feet

Step-by-step explanation:

It is very long process

Answer:

A) 8/ 3(x-5)

Step-by-step explanation:

1) 2x+14/x^2-25 • 8x+40/6x+42=2(x+7)/ (x-5)(x+5) * 8(x+5)/ 6(x+7)

2)

after expressing every part of every fraction as product you should reduce the fraction. You can get rid of common divisor (x+5) in the denominator of the first fraction and the numerator of the second one and the common divisor (x+7) for the numerator of the first fraction and denominatorof the second one.

Then you'll get 2*8/ 6(x+5)   reduce it again, 2 and 6 have the common divisor 2

Then 8/3(x-5) is the correct answer

Answer:

Megan’s at 2.5 inches per week

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Answer:

Step-by-step ex0.72

Answer:

A. Men had less distress from nausea on average than women but we can not determine if this is a significant difference.

Step-by-step explanation:

Working based on the information given, the mean values of each group with with men having an average score of 1.02 and women have an average of 1.79 this reveals that distressing nausea on average is higher in women than in men . However, to test if there is a significant difference would be challenging as the information given isn't enough to make proceed with the test as the standard deviations of the two groups aren't given and no accompanying sample data is given.