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Andru [333]
3 years ago
12

A manufacturer is concerned over the diameter of rubber balls being produced. Too large and they don’t work for certain applicat

ions. Too small and they are subject to little children accidentally choking on them. If the balls are normally distributed with a mean diameter of 1.45 and a standard deviation of .08, what is the value of c , such that 98% of the balls fall between the limits of 1.45 +/- c?
Mathematics
1 answer:
UkoKoshka [18]3 years ago
3 0

Answer:

c = 0.1864

Step-by-step explanation:

To find c, we need to find the z-score where:

P(-z<Z<z) = 98%

Based on the properties of the normal distribution we can said that

P(-z<Z<z) = 98% is equal to:

2*P(Z>z) = 2%    or  P(Z>z)= 1%

So, using the standard normal table, the z score is 2.33

Then, we can calculated the value of c as:

c=z*s

Where s is the standard deviation and z is the z-score. Replacing the values, we get:

c = 2.33*0.08

c = 0.1864

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If Rick wishes to reduce his BMI to 27, he needs to eat fewer kcalories than he expends. For an adolescent who carries excess fa
Nesterboy [21]

This question is not complete because his height and age was not indicated in the above question.

Complete Question:

Rick is a healthy 19-year-old college student who is 70 inches tall and weighs 205 pounds. He has decided to "get a six-pack" over the summer with a diet and exercise program. As part of his new plan, he has stopped drinking soda and is eating more salads in addition to his usual diet. Besides of these changes, he is unclear on how to proceed to reach his fitness goal. Rick's mother wants to make sure his approach will not interfere with his normal growth and development, and has asked him to seek reliable information to help him make a reasonable plan.

If Rick wishes to reduce his BMI to 27, he needs to eat fewer kcalories than he expends. For an adolescent who carries excess fat, the recommended maximal weight loss is one pound per week. Since there are 3500 kcalories in a pound of body fat, a deficit of 3500 kcalories for the week or 500 kcalories per day would be required. Calculate the maximum number of kcalories Rick can consume per day to achieve a weight loss of one pound per week. Assume that his weight is 205 pounds and that his physical activity factor is "low active."

Answer:

2696 kilocalories

Step-by-step explanation:

STEP 1

First we need to calculate Rick's Basal Metabolic Rate( BMR)

Weight in pounds = 205 pounds

Age = 19 years

Height in inches = 70 inches

The formula for calculating Basal Metabolic Rate for men =

BMR = 66.47 + ( 6.24 × weight in pounds ) + ( 12.7 × height in inches ) − ( 6.755 × age in years )

BMR = 66.47 + ( 6.24 × 205 ) + ( 12.7 × 70 ) − ( 6.755 × 19)

= 2247

STEP 2:

The next step is to calculate the maximum amount kilocalories per day Rick should consume

Formula is given as :

For a person with physical activity factor of a ' low Active'

Maximum amount of kilocalories to consume per day = Basal Metabolic rate × 1.2

= 2247 × 1.2

= 2696 Kilocalories (kcalories) per day

6 0
3 years ago
Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean
Mamont248 [21]

Answer:

a) The percentage of snails that take more than 60 hours to finish is 4.75%.

b) The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c) The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d) 0% probability that a randomly-chosen snail will take more than 76 hours to finish

e) To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f) The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

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 = 50, \sigma = 6

a. The percentage of snails that take more than 60 hours to finish is

This is 1 subtracted by the pvalue of Z when X = 60.

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

Z = \frac{60 - 50}{6}

Z = 1.67

Z = 1.67 has a pvalue 0.9525

1 - 0.9525 = 0.0475

The percentage of snails that take more than 60 hours to finish is 4.75%.

b. The relative frequency of snails that take less than 60 hours to finish is

This is the pvalue of Z when X = 60.

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

Z = \frac{60 - 50}{6}

Z = 1.67

Z = 1.67 has a pvalue 0.9525

The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c. The proportion of snails that take between 60 and 67 hours to finish is

This is the pvalue of Z when X = 67 subtracted by the pvalue of Z when X = 60.

X = 67

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

Z = \frac{67 - 50}{6}

Z = 2.83

Z = 2.83 has a pvalue 0.9977

X = 60

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

Z = \frac{60 - 50}{6}

Z = 1.67

Z = 1.67 has a pvalue 0.9525

0.9977 - 0.9525 = 0.0452

The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d. The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 1 subtracted by the pvalue of Z when X = 76.

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

Z = \frac{76 - 50}{6}

Z = 4.33

Z = 4.33 has a pvalue of 1

1 - 1 = 0

0% probability that a randomly-chosen snail will take more than 76 hours to finish

e. To be among the 10% fastest snails, a snail must finish in at most hours.

At most the 10th percentile, which is the value of X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

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

-1.28 = \frac{X - 50}{6}

X - 50 = -1.28*6

X = 42.32

To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f. The most typical 80% of snails take between and hours to finish.

From the 50 - 80/2 = 10th percentile to the 50 + 80/2 = 90th percentile.

10th percentile

value of X when Z has a pvalue of 0.1. So X when Z = -1.28.

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

-1.28 = \frac{X - 50}{6}

X - 50 = -1.28*6

X = 42.32

90th percentile.

value of X when Z has a pvalue of 0.9. So X when Z = 1.28

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

1.28 = \frac{X - 50}{6}

X - 50 = 1.28*6

X = 57.68

The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

5 0
3 years ago
Find the value of x in the<br> isosceles triangle shown below.<br> Help plzzzz
zysi [14]

Answer:

x =2

Option D is correct !!!

Step-by-step explanation:

By Pythagoras theorem we can solve this

we know that:

{x}^{2}  +  {4}^{2}  =  \sqrt{20} {}^{2}

{x}^{2}  =  \sqrt{20} ^{2}   - 16

{x}^{2} = 20 - 16

x = 2

6 0
3 years ago
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Anastasy [175]

Answer:

8

Step-by-step explanation:

3(2)-2(1-2) / 3-2

6+2 / 1

8/1

= 8

5 0
3 years ago
Read 2 more answers
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