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3 years ago

Simplify 14.2 + 6.5(p + 3q) – 4.05q

Mathematics

2 answers:

abruzzese [7]3 years ago

8 0

Step-by-step explanation:

14.2 + 6.5p+ 19.5q- 4.05q

6.5p + 15.45q + 14.2

earnstyle [38]3 years ago

7 0

The answer to your question is C.

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Not sure if u can do it without a protractor but please help

lana [24]

Answer:

i dont know the other questions but i do know the last one and the answer is rhombus.

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3 years ago

Which is equivalent to 4y-2x=16

ad-work [718]

X-intercept. y-intercept
4*0-2x=16. 4y-2*0=16
-2x+0=16. 4y-0=16
-0. -0. +0 +0
-2x. /-2 = 16/-2 4y/4 =16/4
x = -8. y = 4
(-8,0) (0,4)

4 0

3 years ago

How do I solve the absolute value of -5x=45

GREYUIT [131]

Well, You divide both sides by -5.

−5x/−5=45/−5

x=−9

Answer:

x=−9

7 0

3 years ago

The mean amount of money spent on lunch per week for a sample of 100 students is $21. If the margin of error for the population

Digiron [165]

From the given mean and margin of error, the 99% confidence interval for the mean amount of money spent on lunch per week for all students is:

[$19.5, $22.5].

<h3>How to calculate a confidence interval given the sample mean and the margin of error?</h3>

The confidence interval is given by the sample mean plus/minus the margin of error, hence:

The lower bound is the sample mean subtracted by the margin of error.

The upper bound is the sample mean added to the margin of error.

For this problem, we have that:

The sample mean is of $21.

The margin of error is of $1.50.

Hence the bounds are given as follows:

Lower bound: 21 - 1.50 = $19.50.

Upper bound: 21 + 1.50 = $22.50.

Hence the interval is [$19.50, $22.50].

More can be learned about confidence intervals at brainly.com/question/25890103

#SPJ1

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1 year ago

According to survey data, the distribution of arm spans for males is approximately Normal with a mean of 71.4 inches and a stan

Gekata [30.6K]

Answer:

a) 89.97% of men have arm spans between 66 and 76 inches.

b) The z-score for this person's arm span is 5.68. 0% of males have an arm span at least as long as this person

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 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.

Mean of 71.4 inches and a standard deviation of 3.1 inches.

This means that $\mu = 71.4, \sigma = 3.1$

a. What percentage of men have arm spans between 66 and 76 inches?

The proportion is the pvalue of Z when X = 76 subtracted by the pvalue of Z when X = 66. The percentage is the proportion multiplied by 100.

X = 76

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

$Z = \frac{76 - 71.4}{3.1}$

$Z = 1.48$

$Z = 1.48$ has a pvalue of 0.9306

X = 66

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

$Z = \frac{66 - 71.4}{3.1}$

$Z = -1.74$

$Z = -1.74$ has a pvalue of 0.0409

0.9306 - 0.0409 = 0.8997

0.8997*100% = 89.97%

89.97% of men have arm spans between 66 and 76 inches.

b. A particular professional basketball player has an arm span of almost 89 inches. Find the z-score for this person's arm span. What percentage of males have an arm span at least as long as this person?

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

$Z = \frac{89 - 71.4}{3.1}$

$Z = 5.68$

The z-score for this person's arm span is 5.68.

Z = 5.68 has a pvalue of 1

1 - 1 = 0

0% of males have an arm span at least as long as this person

8 0

3 years ago