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Over [174]
2 years ago
15

Washington is 44 miles south of Lawrence. and midland is 63 miles east of Lawrence. how far apart are Washington and midland?

Mathematics
1 answer:
gregori [183]2 years ago
3 0
The answer is 21 miles
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According to a human modeling​ project, the distribution of foot lengths of women is approximately Normal with a mean of 23.3 ce
Yakvenalex [24]

Answer:

26.11% of women in the United States will wear a size 6 or​ smaller

Step-by-step explanation:

Problems of normally distributed samples are 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:

\mu = 23.3, \sigma = 1.4

In the United​ States, a​ woman's shoe size of 6 fits feet that are 22.4 centimeters long. What percentage of women in the United States will wear a size 6 or​ smaller?

This is the pvalue of Z when X = 22.4. So

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

Z = \frac{22.4 - 23.3}{1.4}

Z = -0.64

Z = -0.64 has a pvalue of 0.2611

26.11% of women in the United States will wear a size 6 or​ smaller

8 0
2 years ago
Each gallon of paint covers $200$ square feet. I have to paint one side of a wall that is $12$ meters tall and $80$ meters long.
dsp73

Answer:

make those cheeks go in circuler motion

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
Formulate the indicated conclusion in nontechnical terms. Be sure to address the original claim. The foundation chair for a hosp
slavikrds [6]

Answer:

For this case we want to test if the mean number of filled overnight beds is over​ 523. If X represent our random variable "number of filled overnight beds", the system of hypothesis are:

Null Hypothesis: \mu \leq 523

Alternative hypothesis: \mu > 523

And for this case after conduct the test is FAIL to reject the null hypothesis. So then we can conclude that the claim that the number of filled overnight beds is over​ 523 is not statistically supported

Step-by-step explanation:

Previous concepts

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".

Solution to the problem

For this case we want to test if the mean number of filled overnight beds is over​ 523. If X represent our random variable "number of filled overnight beds", the system of hypothesis are:

Null Hypothesis: \mu \leq 523

Alternative hypothesis: \mu > 523

And for this case after conduct the test is FAIL to reject the null hypothesis. So then we can conclude that the claim that the number of filled overnight beds is over​ 523 is not statistically supported

6 0
2 years ago
At the amusement park, Laura paid $6.00 for a small cotton candy. Her older brother works at the park, and he told
BabaBlast [244]

Answer:

yes

The price of marked up candy is $6.00

perentage of mark up is 300%

let the marked up amount be represented as x

let 300% mark up be represented as 3x

the equatiopn for finding the price before markup

can be represented as 3x-x=$6.

                                    2x=$6

                                    x=$3.

8 0
2 years ago
What is 4.66666666667 as a fraction
ss7ja [257]

Step-by-step explanation:

466666666667/100000000000

7 0
2 years ago
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