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Anna71 [15]
3 years ago
8

You spend 1/2 of your allowance on Monday. On Tuesday, you spend 1/2 of the amount you have left. On Wednesday, you spend 1/2 of

what remains.
What fraction of your allowance do you have left?
Mathematics
1 answer:
dimulka [17.4K]3 years ago
4 0
Given that the allowance is variable. It varies in a week.
Let X = be your allowance.

Monday to Wednesday, you already incurred
Monday = X / 2
Tuesday = Monday / 2
Wednesday = Tuesday / 2

The money left you have will be
Money left = X - (Monday + Tuesday + Wednesday)

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Make x the subject of the formula if a/b-x=b/a+x​
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3 years ago
SOMEONE HELP ME ON THESE QUESTIONS
dem82 [27]

1.)  The sum(addition) of 21 and 5 times(multiplication) a number f is(=) 61.

f = unknown number/variable     [So 21 plus 5f(5 times f) equals 61]

21 + 5f = 61   [21(one-time) + 5f(number x variable) = 61(total)]

2.)  Seventeen more(addition) than seven times(multiplication) a number j is(=) 87.

j = unknown number/variable    [So 17 plus 7j(7 times j) equals 87]

17 + 7j = 87  

3.)   n = number of calls

18 + 0.05n = 50.50  

[Company charges $18 plus five cents per call(n), and the total charge was $50.50]

4.)     s = the number of students

40 + 30s = 220

[Tutor charges $40 plus $30 per student(s), and the total charge was $220]

4 0
3 years ago
The average length of a field goal in the National Football League is 38.4 yards, and the s. d. is 5.4 yards. Suppose a typical
Tasya [4]

Answer:

a) The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b) 0.25% probability that his average kicks is less than 36 yards

c) 0.11% probability that his average kicks is more than 41 yards

d-a) The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 1.08

d-b) 1.32% probability that his average kicks is less than 36 yards

d-c) 0.80% probability that his average kicks is more than 41 yards

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 38.4, \sigma = 5.4, n = 40, s = \frac{5.4}{\sqrt{40}} = 0.8538

a. What is the distribution of the sample mean? Why?

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 0.8538.

b. What is the probability that his average kicks is less than 36 yards?

This is the pvalue of Z when X = 36. So

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

By the Central Limit Theorem

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

Z = \frac{36 - 38.4}{0.8538}

Z = -2.81

Z = -2.81 has a pvalue of 0.0025

0.25% probability that his average kicks is less than 36 yards

c. What is the probability that his average kicks is more than 41 yards?

This is 1 subtracted by the pvalue of Z when X = 41. So

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

Z = \frac{41 - 38.4}{0.8538}

Z = 3.05

Z = 3.05 has a pvalue of 0.9989

1 - 0.9989 = 0.0011

0.11% probability that his average kicks is more than 41 yards

d. If the sample size is 25 in the above problem, what will be your answer to part (a) , (b)and (c)?

Now n = 25, s = \frac{5.4}{\sqrt{25}} = 1.08

So

a)

The sample is larger than 30, so, by the Central Limit Theorem, the distribution of the sample means will be normally distributed with mean 38.4 and standard deviation 1.08

b)

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

Z = \frac{36 - 38.4}{1.08}

Z = -2.22

Z = -2.22 has a pvalue of 0.0132

1.32% probability that his average kicks is less than 36 yards

c)

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

Z = \frac{41 - 38.4}{1.08}

Z = 2.41

Z = 2.41 has a pvalue of 0.9920

1 - 0.9920 = 0.0080

0.80% probability that his average kicks is more than 41 yards

4 0
3 years ago
Need help ASAP!!!!!!!!!!!!!!!!!
rosijanka [135]

Answer:

A 97

Step-by-step explanation:

Adds to 180 degrees. 83 + 97 = 180

8 0
2 years ago
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