Ask question

Ask question

All categories

3 years ago

13

Jason's meeting starts at 3 p.m. and is 270 miles away. He estimates that he can average 60 miles per hour. What time should he

leave if he wants to arrive on time?

1 answer:

Maurinko [17]3 years ago

4 0

Answer:

He should leave by 10:30 a.m

Step-by-step explanation:

If he drives 60 miles per hr it will take him 270÷60=4.5 (4 n half) hrs to reach there.

If he want to be there at 3p.m he should leave by 10:30 a.m

You might be interested in

Solve for X<br><br> PLEASE HELP ME

guapka [62]

Answer:

x = 10°

Step-by-step explanation:

Step 1:

92° + x° + 78° = 180° Supplementary Angles

Step 2:

x° + 170° = 180° Combine Like Terms

Step 3:

x° = 180° - 170° Subtract 170° on both sides

Answer:

x = 10°

Hope This Helps :)

4 0

3 years ago

How to factor 5y+11y+2

aksik [14]

16y+2 is your answer

7 0

3 years ago

If<br>find the values of x and y.

uysha [10]

5-2=3 so x is equal to 3
-3+10 is equal to 7 so y is equal to 10
With this we can conclude that x=3 and y=10
I hope that helps :)

8 0

3 years ago

In the diagram, which two angles are alternate interior angles with angle 14?

mars1129 [50]

The answer would be A.) angle 4 and angle 12

4 0

4 years ago

Assume that weights of adult females are normally distributed with a mean of 79 kg and a standard deviation of 22 kg. What perce

LenKa [72]

Answer:

14.28% of individual adult females have weights between 75 kg and 83 kg.

92.82% of the sample means are between 75 kg and 83 kg.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

Assume that weights of adult females are normally distributed with a mean of 79 kg and a standard deviation of 22 kg. This means that $\mu = 79, \sigma = 22$ .

What percentage of individual adult females have weights between 75 kg and 83 kg?

This percentage is the pvalue of Z when $X = 83$ subtracted by the pvalue of Z when $X = 75$ . So:

X = 83

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{83 - 79}{22}$

$Z = 0.18$

$Z = 0.18$ has a pvalue of 0.5714.

X = 75

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{75- 79}{22}$

$Z = -0.18$

$Z = -0.18$ has a pvalue of 0.4286.

This means that 0.5714-0.4286 = 0.1428 = 14.28% of individual adult females have weights between 75 kg and 83 kg.

If samples of 100 adult females are randomly selected and the mean weight is computed for each sample, what percentage of the sample means are between 75 kg and 83 kg?

Now we use the Central Limit THeorem, when $n = 100$ . So $s = \frac{22}{\sqrt{100}} = 2.2.$

X = 83

$Z = \frac{X - \mu}{s}$

$Z = \frac{83 - 79}{2.2}$

$Z = 1.8$

$Z = 1.8$ has a pvalue of 0.9641.

X = 75

$Z = \frac{X - \mu}{s}$

$Z = \frac{75-79}{2.2}$

$Z = -1.8$

$Z = -1.8$ has a pvalue of 0.0359.

This means that 0.9641-0.0359 = 0.9282 = 92.82% of the sample means are between 75 kg and 83 kg.

8 0

4 years ago