A car with a mass of 500 kg is accelerating at 10 m/s². What is the force being applied the car’s engine?

It is generally claimed that the average body temperature for healthy human adults is 98.6°F. To test this claim a simple random

Rzqust [24]

Answer:

Explained below.

Step-by-step explanation:

The information provided is as follows:

$\mu=98.6^{o}F\\\bar x=98.1^{o}F\\s=0.9^{o}F\\n=9$

(1)

A single mean test is to be performed in this case.

As the population standard deviation is not provided, a one-sample t-test will be used.

The correct option is b.

(2)

The null hypothesis is:

H₀: The average temperature in the population is 98.6°F, i.e. μ = 98.6°F.

The correct option is b.

(3)

The alternative hypothesis is:

Hₐ: The average temperature in the population is less than 98.6°F, i.e. μ < 98.6°F.

The correct option is c.

(4)

The standard deviation of the sample mean is as follows:

$SD_{\bar x}=\frac{s}{\sqrt{n}}=\frac{0.9}{\sqrt{9}}=0.3^{o}F$

Thus, the value of SD is 0.3°F.

(5)

Compute the value of test statistic as follows:

$t=\frac{\bar x-\mu}{SD_{\bar x}}$

$=\frac{98.1-98.6}{0.3}\\\\=-1.666666667\\\\\approx -1.67$

Thus, the value of test statistic is -1.67.

(6)

The degrees of freedom of the test are:

df = n - 1

= 9 - 1

= 8

Thus, the degrees of freedom of the test is 8.

8 0

3 years ago

Can someone explain to me how to answer questions like this? I always get confused!

seropon [69]

Answer:

(b) 5k+1

Step-by-step explanation:

https://www.symbolab.com/solver/step-by-step/simplify%20%5Cfrac%7B30k%5E%7B2%7D%2B6k%7D%7B5k%2Bk%7D

5 0

3 years ago

Quadrilateral abcd is a trapezoid with what must be shown to prove that abcd is an isosceles trapezoid?

monitta

Do you have a picture

4 0

3 years ago

According to a study done by Wakefield Research, the proportion of Americans who can order a meal in a foreign language is 0.47.

UNO [17]

Answer:

Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.

Step-by-step explanation:

We are given that according to a study done by Wake field Research, the proportion of Americans who can order a meal in a foreign language is 0.47.

Suppose a random sample of 200 Americans is asked to disclose whether they can order a meal in a foreign language.

Let $\hat p$ = sample proportion of Americans who can order a meal in a foreign language

The z-score probability distribution for sample proportion is given by;

Z = $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion

p = population proportion of Americans who can order a meal in a foreign language = 0.47

n = sample of Americans = 200

Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is given by = P( $\hat p$ > 0.50)

P( $\hat p$ > 0.06) = P( $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ > $\frac{0.5-0.47}{\sqrt{\frac{0.5(1-0.5)}{200} } }$ ) = P(Z > 0.85) = 1 - P(Z $\leq$ 0.85)

= 1 - 0.80234 = 0.19766

Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.85 in the z table which has an area of 0.80234.

Therefore, probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.

4 0

3 years ago

Question One

Alex

Answer:

Step-by-step explanation:

Set up proportions. Make sure length and widths line up!

$\frac{12}{5} =\frac{x}{8}$