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Orlov [11]
3 years ago
15

Help me please with these questions

Mathematics
2 answers:
Setler79 [48]3 years ago
8 0
17. B $3.00
18.D 35/36
19.D 89.775
Lapatulllka [165]3 years ago
3 0

Answer:

17) option b

18) option b

19) option d

Step-by-step explanation:

17)

Cost of 25 burgers = $75

Therefore , cost of 1 burger:

                                          \frac{75}{25} = 3

18)

Area of  rectangle = length x breadth

                 \frac{7}{6} = length \times \frac{5}{6}\\\\\frac{7}{6} \times \frac{6}{5} = length \\\\\frac{7}{5} = length \\\\Length = \frac{7}{5}  \ km

19)

                   14.25

                 x    6.3

             _________

                   4275

            +    8550              

             __________

                89.775

            ___________

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In a survey, 24 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped
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Answer:

The 98% confidence interval for the mean amount spent on their child's last birthday gift is between $40.98 and $43.02.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 24 - 1 = 23

98% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 23 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.98}{2} = 0.99. So we have T = 2.5

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 2.5\frac{2}{\sqrt{24}} = 1.02

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 42 - 1.02 = $40.98.

The upper end of the interval is the sample mean added to M. So it is 42 + 1.02 = $43.02.

The 98% confidence interval for the mean amount spent on their child's last birthday gift is between $40.98 and $43.02.

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3 years ago
A teacher interested in determining the effect of a new computer program on learning to read conducted a study. One hundred stud
erastova [34]

Answer:

This is an experiment because a treatment was applied to a group.

Step-by-step explanation:

This is because it cannot be a survey since the children are taking a test, which has a hypothesis. Hope you get this right. I took the test and got right.

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Step-by-step explanation:

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8.52 The heights of 2-year-old children are normally distributed with a mean of 32 inches and a standard deviation of 1.5 inches
sweet [91]

Answer:

Heights of 29.5 and below could be a problem.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

The heights of 2-year-old children are normally distributed with a mean of 32 inches and a standard deviation of 1.5 inches.

This means that \mu = 32, \sigma = 1.5

There may be a problem when a child is in the top or bottom 5% of heights. Determine the heights of 2-year-old children that could be a problem.

Heights at the 5th percentile and below. The 5th percentile is X when Z has a p-value of 0.05, so X when Z = -1.645. Thus

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

-1.645 = \frac{X - 32}{1.5}

X - 32 = -1.645*1.5

X = 29.5

Heights of 29.5 and below could be a problem.

4 0
3 years ago
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