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3 years ago

A driving moving at 22m/s stops the car in 2s. find the stopping distance

Mathematics

1 answer:

Kaylis [27]3 years ago

8 0

Step-by-step explanation:

Hey there!

Here,

Speed = 22m/s

Time = 2 s.

Distance = ?

We have;

$d = s \times t$

$d = \frac{22m}{s} \times 2 \: s$

Cancel "s" and multiply 22 to 2 ,we will get;

Distance = 44m.

Therefore, distance = 22m.

Hope it helps..

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Which of the following equations represents the line with a slope of -7/4 and a y-intercept of -11?

andreev551 [17]

Y= -7/4x - 11
My answer has to be 20 characters longkdmmwmmckkdmsnskckd

4 0

3 years ago

A study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school

nekit [7.7K]

Answer:

We conclude that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids.

Step-by-step explanation:

We are given that a study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois.

Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids.

Let $p_1$ = proportion of Illinois high school freshmen who have used anabolic steroids.

$p_2$ = proportion of Illinois high school seniors who have used anabolic steroids.

SO, Null Hypothesis, $H_0$ : $p_1=p_2$ {means that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids}

Alternate Hypothesis, $H_A$ : $p_1\neq p_2$ {means that there is a significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids}

The test statistics that would be used here Two-sample z test for proportions;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of high school freshmen who have used anabolic steroids = $\frac{34}{1679}$ = 0.0203

$\hat p_2$ = sample proportion of high school seniors who have used anabolic steroids = $\frac{24}{1366}$ = 0.0176

$n_1$ = sample of high school freshmen = 1679

$n_2$ = sample of high school seniors = 1366

So, the test statistics = $\frac{(0.0203-0.0176)-(0)}{\sqrt{\frac{0.0203(1-0.0203)}{1679}+\frac{0.0176(1-0.0176)}{1366} } }$

= 0.545

The value of z test statistics is 0.545.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids.

3 0

3 years ago

Gcf using prime factorization of 52 and 65

Leto [7]

25: 2 × 2 × 13 65: 5 × 13

3 0

3 years ago

What is the smallest and largest number that round to 170,000

Maksim231197 [3]

Assuming you're rounding to the nearest ten-thousand, Smallest is 165,000and largest is 174,000

8 0

3 years ago

The circumference of a sphere was measured to be 76 cm with a possible error of 0.5 cm. (a) Use differentials to estimate the ma

andreev551 [17]

The maximum error in the calculated surface area is 24.19cm² and the relative error is 0.0132.

Given that the circumference of a sphere is 76cm and error is 0.5cm.

The formula of the surface area of a sphere is A=4πr².

Differentiate both sides with respect to r and get

dA÷dr=2×4πr

dA÷dr=8πr

dA=8πr×dr

The circumference of a sphere is C=2πr.

From above the find the value of r is

r=C÷(2π)

By using the error in circumference relation to error in radius by:

Differentiate both sides with respect to r as

dr÷dr=dC÷(2πdr)

1=dC÷(2πdr)

dr=dC÷(2π)

The maximum error in surface area is simplified as:

Substitute the value of dr in dA as

dA=8πr×(dC÷(2π))

Cancel π from both numerator and denominator and simplify it

dA=4rdC

Substitute the value of r=C÷(2π) in above and get

dA=4dC×(C÷2π)

dA=(2CdC)÷π

Here, C=76cm and dC=0.5cm.

Substitute this in above as

dA=(2×76×0.5)÷π

dA=76÷π

dA=24.19cm².

Find relative error as the relative error is between the value of the Area and the maximum error, therefore:

$\begin{aligned}\frac{dA}{A}&=\frac{8\pi rdr}{4\pi r^2}\\ \frac{dA}{A}&=\frac{2dr}{r}\end$

As above its found that r=C÷(2π) and r=dC÷(2π).

Substitute this in the above

$\begin{aligned}\frac{dA}{A}&=\frac{\frac{2dC}{2\pi}}{\frac{C}{2\pi}}\\ &=\frac{2dC}{C}\\ &=\frac{2\times 0.5}{76}\\ &=0.0132\end$

Hence, the maximum error in the calculated surface area with the circumference of a sphere was measured to be 76 cm with a possible error of 0.5 cm is 24.19cm² and the relative error is 0.0132.

Learn about relative error from here brainly.com/question/13106593

#SPJ4

3 0

2 years ago