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garik1379 [7]
3 years ago
5

Please solve this !!! ASAP !!!

Mathematics
1 answer:
ivolga24 [154]3 years ago
5 0

the answer is

48 2/9 ft.

I did the equation for it

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The breaking strengths of cables produced by a certain manufacturer have a mean, , of pounds, and a standard deviation of pounds
olchik [2.2K]

The question is incomplete. The complete question is :

The breaking strengths of cables produced by a certain manufacturer have a mean of 1900 pounds, and a standard deviation of 65 pounds. It is claimed that an improvement in the manufacturing process has increased the mean breaking strength. To evaluate this claim, 150 newly manufactured cables are randomly chosen and tested, and their mean breaking strength is found to be 1902 pounds. Assume that the population is normally distributed. Can we support, at the 0.01 level of significance, the claim that the mean breaking strength has increased?

Solution :

Given data :

Mean, μ = 1900

Standard deviation, σ = 65

Sample size, n = 150

Sample mean, $\overline x$ = 1902

Level of significance = 0.01

The hypothesis are :

$H_0 : \mu = 1900$

$H_1 : \mu > 1900$

Test statics :

We use the z test as the sample size is large and we know the population standard deviation.

$z=\frac{\overline x - \mu}{\sigma / \sqrt{n}}$

$z=\frac{1902-1900}{65 / \sqrt{150}}$

$z=\frac{2}{5.30723}$

$z=0.38$

Finding the p-value:

P-value = P(Z > z)

             = P(Z > 0.38)

             = 1 - P(Z < 0.38)

From the z table. we get

P(Z < 0.38) = 0.6480

Therefore,

P-value = 1 - P(Z < 0.38)

            = 1 - 0.6480

             = 0.3520

Decision :

If the p value is less than 0.01, then we reject the H_0, otherwise we fail to reject  H_0.

Since the value of p = 0.3520 > 0.01, the level of significance, then we fail to reject  H_0.

Conclusion :

At a significance level of 0.01, we have no sufficient evidence to support that the mean breaking strength has increased.

4 0
2 years ago
5. Sarah made 17 out of 25 free throws during practice. What percentage of free throws did she NOT make? A. 68% B. 58% C. 32% D.
Elanso [62]

Answer:

C. 32%

Step-by-step explanation:

The percent 17 is of 25.

17 of 25 can be written as:

17/25

To find percentage, we need to find an equivalent fraction with denominator 100. Multiply both numerator & denominator by 100

17/25 * 100/100

= (17 * 100/25) * 1/100 = 68/100

Therefore, the answer is 68%

If Sara made 68% of her free shows, that means she didn't make 32% of her free throws (100 - 68 = 32).

5 0
2 years ago
Help pls thank you!!!!!!
jasenka [17]

Answer:

relative minimum would be your awnser

4 0
3 years ago
Read 2 more answers
1 /x^−1 in simplest rational form
Allushta [10]
Here is a link to help you find the answer:
https://www.khanacademy.org/math/algebra2/rational-expressions-equations-and-functions/simplify-rati...

Hope this helps!
3 0
3 years ago
Genetic Defects Data indicate that a particular genetic defect occurs in of every children. The records of a medical clinic show
Dovator [93]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability that there exist 60 or more defected children is P(x \ge 60)=0.0901

Looking at the value for this probability we see that it is not so small to the point that the observation of this kind would be a rare occurrence

Step-by-step explanation:

From the question we are told that

        in every 1000 children a particular genetic defect occurs to 1

        The number of sample selected is n= 50,000

The probability of observing the defect is mathematically evaluated as

              p = \frac{1}{1000}

                 = 0.001

The probability of not observing the defect is mathematically evaluated as

            q = 1-p

               = 1-0.001

               = 0.999

The mean of this probability is mathematically represented as

                 \mu = np

Substituting values

                \mu = 50000*0.001

                    = 50

The standard deviation of this probability is mathematically represented as

   \sigma = \sqrt{npq}

Substituting values

      = \sqrt{50000 * 0.001 * 0.999}

     = \sqrt{49.95}

     = 7.07

 the probability of detecting  x  \ge60 defects can be represented in as  normal distribution like

       P(x \ge 60)

in standardizing the normal distribution the normal area used to approximate P(x \ge 60) is the right of 59.5 instead of 60 because  x= 60 is part of the observation

The z -score is obtained mathematically as

                z = \frac{x-\mu }{\sigma }

                   = \frac{59.5 - 50 }{7.07}

                  =1.34

The area to the left of z = 1.35 on the standardized normal distribution curve is 0.9099 obtained from the z-table shown z value to the left of the standardized normal curve

Note: We are looking for the area to the right i.e 60 or more

  The total area under the curve is 1

So

    P(x \ge 60) \approx P(z > 1.34)

                     = 1-P(z \le 1.34)

                    =1-0.9099

                  =0.0901

             

   

3 0
3 years ago
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