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3 years ago

Pls help me I need this now pls

Mathematics

2 answers:

Anarel [89]3 years ago

7 0

Answer: C

Step-by-step explanation:

LenaWriter [7]3 years ago

4 0

Answer:

450 cm

Step-by-step explanation:

25 on both sides 2x25

4 squares = 25x4

4 sides for 4 squares = 100x4

400+50

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Find the sum of the geometric series: 6+12+24+48+...+12,288. Show your work

slava [35]

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3 years ago

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Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given question.

inn [45]

Answer:

Step-by-step explanation:

(a) MEAN = £fx/ fx (The sum of the all jerseys number/ number of jerseys)

= 680/11

= 61.8

(b) MEDIAN = The middle number when all the items are arranged either in ascending order or descending order

= 6, 14, 50, 51, 54, 68, 77, 78, 89, 94, 99

= 68

(Strike out the numbers, the middle number left is the median)

From those items, THERE IS NO MODE.

All the items (numbers) appeared once

(d) MID-RANGE:

This is the sum of the least valued number and the greatest valued number and dividee by two

Mid-range = (99 + 6) / 2

= 105/2

= 52.5

(e). The results tells us that:

The mean and median give two different interpretations of the average (or typical) jersey number, while the midrange shows the spread of possible jersey numbers

Also, pls note that the median implies the measure of the middle of the sets of jerseys number. And the mean tells us the distribution of the sets of jerseys number.

Hope this helped!

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3 years ago

Which of the following represents<br> this statement? |3x – 41 < 2

agasfer [191]

<h2>Answer:</h2><h2>D is the correct answer </h2><h2></h2><h2>Hope this helps!!</h2>

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3 years ago

A half-century ago, the mean height of women in a particular country in their 20s was 64.7 inches. Assume that the heights of

Ainat [17]

Answer:

99.5% of all samples of 21 of today's women in their 20's have mean heights of at least 65.86 inches.

Step-by-step explanation:

We are given that a half-century ago, the mean height of women in a particular country in their 20's was 64.7 inches. Assume that the heights of today's women in their 20's are approximately normally distributed with a standard deviation of 2.07 inches.

Also, a samples of 21 of today's women in their 20's have been taken.

<u><em>Let </em></u> $\bar X$ <u><em> = sample mean heights</em></u>

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean height of women = 64.7 inches

$\sigma$ = standard deviation = 2.07 inches

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, Probability that the sample of 21 of today's women in their 20's have mean heights of at least 65.86 inches is given by = P( $\bar X$ $\geq$ 65.86 inches)

P( $\bar X$ $\geq$ 65.86 inches) = P( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ $\geq$ $\frac{65.86-64.7}{\frac{2.07}{\sqrt{21} } }$ ) = P(Z $\geq$ -2.57) = P(Z $\leq$ 2.57)