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kolezko [41]
2 years ago
7

Jacob spent half of his weekly allowance at the movies. To earn more money his parents let him wash the car for $7. What is his

weekly allowance (w) if he ended up with $13? The variable should always go in the first term.
Mathematics
1 answer:
sweet [91]2 years ago
5 0

Answer:

6 dollers couse 13 minuse 7 equels to 6

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D -7π4

Step-by-step explanation:

Just took the test my dude :D

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What is the solution to this equation x+6=24
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x=18 if im not mistaken that was pretty easY

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Pablo uses 4 centimeters of tape for every present he wraps . How much tape will Pablo need in all if he has to wrap 9 presents?
OLga [1]

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36 centimeters of tape

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3 years ago
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Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distribute
balu736 [363]

Answer:

The area of the shaded region is 0.9082.

Step-by-step explanation:

Problems of normally distributed samples can be 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, which is the area of the shaded region, 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 = 100, \sigma = 15, X = 120

Area of the shaded region:

pvalue of Z

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

Z = \frac{120 - 100}{15}

Z = 1.33

Z = 1.33 has a pvalue of 0.9082

So

The area of the shaded region is 0.9082.

4 0
3 years ago
A random sample of 15 employees was selected. The average age in the sample was 31 years with a variance of 49 years. Assuming a
vova2212 [387]

Answer:

Assuming ages are normally distributed, the 98% confidence interval for the population average age is [26.3, 35.7].

Step-by-step explanation:

We have to construct a 98% confidence interval for the mean.

The information we have is:

- Sample mean: 31

- Variance: 49

- Sample size: 15

- The age is normally distributed.

We know that the degrees of freedom are

k=n-1=15-1=14

Then, the t-value for a 98% CI is t=2.625 (according to the t-table).

The standard deviation can be estimated from the variance as:

s=\sqrt {s^2}=\sqrt{49}=7

The margin of error is:

E=t_{14}*s/\sqrt{n}\\\\E=2.625*7/\sqrt{15}=18.375/3.873=4.7

Then, the CI can be constructed as:

M-t*s/\sqrt{n}\leq\mu\leq M+t*s/\sqrt{n}\\\\31-4.7\leq \mu \leq 31+4.7\\\\26.3\leq \mu \leq 35.7

5 0
3 years ago
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