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Sergio039 [100]

3 years ago

10

Seth traveled 1 mile in 15.1 seconds.About how fast does seth travel in miles per hour?

1 answer:

AnnyKZ [126]3 years ago

7 0

Answer

1 mile / 15.1sec * 3,600 sec / hour = 238.4105960265 miles / hour

Step-by-step explanation:

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Helppppppp meee ASAP

balandron [24]

$75|5\\15|5\\.\ 3|3\\.\ 1|\\75=\boxed{5}\cdot\boxed{5}\cdot3\\\\100|5\\.\ 20|5\\.\ \ 4|2\\.\ \ 2|2\\.\ \ 1|\\100=\boxed{5}\cdot\boxed{5}\cdot2\cdot2\\\\175|5\\.\ 35|5\\.\ \ 7|7\\.\ \ 1|\\175=\boxed{5}\cdot\boxed{5}\cdot7\\\\GCF(75;\ 100;\ 175)=\boxed{5}\cdot\boxed{5}=25$

Answer: 25

6 0

3 years ago

Read 2 more answers

The following results come from two independent random samples taken of two populations.

photoshop1234 [79]

Answer:

$(a)\ \bar x_1 - \bar x_2 = 2.0$

$(b)\ CI =(1.0542,2.9458)$

$(c)\ CI = (0.8730,2.1270)$

Step-by-step explanation:

Given

$n_1 = 60$ $n_2 = 35$

$\bar x_1 = 13.6$ $\bar x_2 = 11.6$

$\sigma_1 = 2.1$ $\sigma_2 = 3$

Solving (a): Point estimate of difference of mean

This is calculated as: $\bar x_1 - \bar x_2$

$\bar x_1 - \bar x_2 = 13.6 - 11.6$

$\bar x_1 - \bar x_2 = 2.0$

Solving (b): 90% confidence interval

We have:

$c = 90\%$

$c = 0.90$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.90$

$\alpha = 0.10$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.10/2}$

$z_{\alpha/2} = z_{0.05}$

The z score is:

$z_{\alpha/2} = z_{0.05} =1.645$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.645 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.645 * \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.645 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.645 * \sqrt{0.3306}$

$2.0 \± 0.9458$

Split

$(2.0 - 0.9458) \to (2.0 + 0.9458)$

$(1.0542) \to (2.9458)$

Hence, the 90% confidence interval is:

$CI =(1.0542,2.9458)$

Solving (c): 95% confidence interval

We have:

$c = 95\%$

$c = 0.95$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.95$

$\alpha = 0.05$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.05/2}$

$z_{\alpha/2} = z_{0.025}$

The z score is:

$z_{\alpha/2} = z_{0.025} =1.96$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.96 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.96* \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.96 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.96* \sqrt{0.3306}$

$2.0 \± 1.1270$

Split

$(2.0 - 1.1270) \to (2.0 + 1.1270)$

$(0.8730) \to (2.1270)$

Hence, the 95% confidence interval is:

$CI = (0.8730,2.1270)$

8 0

3 years ago

What early human development was necessary for survival in cold environments?

AlekseyPX

I thought it would be c but try b or a

5 0

3 years ago

Read 2 more answers

Jane is 3 years older than mary, and mary is twice as old as kay. if kay is x years old, how old is jane?

krek1111 [17]

The answer is y = 2x + 3.

x - Kay's age
y - Jane's age
m - Mary's age

<span>Jane is 3 years older than Mary: y = m + 3
</span><span>Mary is twice as old as Kay: m = 2x
</span><span>
y = m + 3
m = 2x

Substitute m in the first equation:
y = 2x + 3
</span>

6 0

3 years ago

Read 2 more answers

It is believed that the average amount of money spent per U.S. household per week on food is about $98, with standard deviation

Makovka662 [10]

Answer:

Test statistic = 2

P-value = 0.0227

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = $98

Sample mean, $\bar{x}$ = $100

Sample size, n = 100

Population standard deviation, σ = $10

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 98\text{ dollars}\\H_A: \mu > 98\text{ dollars}$

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{100 - 98}{\frac{10}{\sqrt{100}} } = 2$

Now, we can calculate the p-value from the normal table

P-value = 0.0227

6 0

4 years ago