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3 years ago

5

Answer all parts in the question below. Be sure to show all your work. Remember NO WORK, NO CREDIT EVEN IF THE ANSWER IS CORRECT

!!!
13. A car travels 30 ⅕ miles in ⅔ of an hour. What is the average speed, in miles per hour, of the car?

1 answer:

liraira [26]3 years ago

3 0

I'm sorry I need the points lol anyways I hope you got the answer to it

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The temperature of coffee served at a restaurant is normally distributed with an average temperature of 160 degrees Fahrenheit a

posledela

Answer:

The 20th percentile of coffee temperature is 155.4532 degrees Fahrenheit.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 160 degrees

Standard Deviation, σ = 5.4 degrees

We are given that the distribution of temperature of coffee is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.2

$P( X < x) = P( z < \displaystyle\frac{x - 160}{5.4})=0.2$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 160}{5.4} = -0.842\\\\x = 155.4532$

Thus,

$P_{20}=155.4532$

The 20th percentile of coffee temperature is 155.4532 degrees Fahrenheit.

7 0

3 years ago

Simplify 1/18 / 1/6

mariarad [96]

The equation in the question is not the same as the one in the picture so here's the answers to both of them.

Q1 - 1/3 or 0.333333

Q2 - 1/108 or 0.009259

8 0

3 years ago

What’s the answer to this?

Scorpion4ik [409]

Answer:

your answers will be options <u>B and E</u>

Step-by-step explanation:

hope this helps have a nice day!!

8 0

2 years ago

web based software company is interested in estimating the proportion of individuals who use the Firefox browser. In a sample of

nata0808 [166]

<h2>Answer with explanation:</h2>

Let $\hat{p}$ denotes the sample proportion.

As per given , we have

n= 200

$\hat{p}=\dfrac{30}{200}=0.15$

Critical value for 99% confidence : $z_{\alpha/2}=2.576$

Confidence interval for population proportion :-

$\hat{p}\pm z_{\alpha/2} \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.15\pm (2.576)(\sqrt{\dfrac{0.15(1-0.15)}{200}})\\\\\approx0.15\pm0.065\\\\=(0.15-0.065,\ 0.15+0.065)\\\\=(0.085,\ 0.215)$

Hence, a 99% confidence interval for the proportion of all individuals that use Firefox: $(0.085,\ 0.215)$

The lower limit on the 99% confidence interval = 0.085

6 0

3 years ago

A conjecture and a portion of the flowchart proof used to prove the conjecture are shown.

vova2212 [387]

3x - 1 = 26

substitution property of equality

3x = 27

answer
3x = 27

3 0

3 years ago

Read 2 more answers