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saveliy_v [14]
2 years ago
9

Please help

Mathematics
2 answers:
Afina-wow [57]2 years ago
8 0

Answer:

Let M represent the amout at any time.

She saves $40 per week so total amount becomes 40x in any specified week.

where x is the number of weeks.

In the start she has $500 in her bank account so we 'll add this amount into the savings per week. So expression becomes

M = 40x+500

Simply put number of week in place of x and find the total amount.

Step-by-step explanation:

Anna007 [38]2 years ago
3 0

Answer:

for c it is no

Step-by-step explanation:

it is no because she deposits 40 a week, in 4 weeks she deposits 40x4=160, which is nowhere close to the 700 she needs. Even if you count the 500 in her account it is still no since it would total up to 660, not 700

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Find all polar coordinates of point P = (6, 31°).
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A quality control manager at an auto plant measures the paint thickness on newly painted cars. A certain part that they paint ha
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Answer:

78.88% probability that the mean thickness in the sample of 100 points is within 0.1 mm of the target value

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 = 2, \sigma = 0.8, n = 100, s = \frac{0.8}{\sqrt{100}} = 0.08

What is the probability that the mean thickness in the sample of 100 points is within 0.1 mm of the target value?

This is the pvalue of Z when X = 2 + 0.1 = 2.1 subtracted by the pvalue of Z when X = 2 - 0.1 = 1.9. So

X = 2.1

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

By the Central Limit Theorem

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

Z = \frac{2.1 - 2}{0.08}

Z = 1.25

Z = 1.25 has a pvalue of 0.8944

X = 1.9

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

Z = \frac{1.9 - 2}{0.08}

Z = -1.25

Z = -1.25 has a pvalue of 0.1056

0.8944 - 0.1056 = 0.7888

78.88% probability that the mean thickness in the sample of 100 points is within 0.1 mm of the target value

8 0
2 years ago
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