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4 years ago

How do you do this (the first one)? i've been sitting trying to do this for literally half an hour.

Mathematics

1 answer:

joja [24]4 years ago

6 0

Spain = 140
Italy = 40
France = 60
U.S.A = 120
total = 18,000,000

Spain : 140/360 = 39%.....0.39(18,000,000) = 7,020,000
Italy : 40/360 = 11%....0.11(18,000,000) = 1,980,000
France : 60/360 = 17%....0.17(18,000,000) = 3,060,000
U.S.A : 120/360 = 33%...0.33(18,000,000) = 5,940,000

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I have to find an equation for a table

Anna35 [415]

Let's say imput is i and that output is o.

The formula is 2 ^ i = o

2^0 = 1

2^1 = 2

2^2 = 4

2^3 = 8

2^4 = 16

2^5 = 32

Notice how the numbers match?

Hope this helped :)

8 0

3 years ago

HELP PLEASE DON'T UNDERSTAND

lana66690 [7]

For #3 the order on the number line from left to right is -2 1/6, -1.62, -1.26, .21, 3/11, 5/3, 2 2/9, 2.375
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8 0

3 years ago

Umm what are the answers for :

navik [9.2K]

Step-by-step explanation:

21% of 36570

21/100 x 36570 = 7679.7

12.2% of 640

12.2/100 x 640 = 78.08

87.5% of 860

87.5/100 x 860 = 752.5

37.5% of 3200

37.5/100 x 3200 = 1200

4 0

3 years ago

Assume that weights of adult females are normally distributed with a mean of 79 kg and a standard deviation of 22 kg. What perce

LenKa [72]

Answer:

14.28% of individual adult females have weights between 75 kg and 83 kg.

92.82% of the sample means are between 75 kg and 83 kg.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

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 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:

Assume that weights of adult females are normally distributed with a mean of 79 kg and a standard deviation of 22 kg. This means that $\mu = 79, \sigma = 22$ .

What percentage of individual adult females have weights between 75 kg and 83 kg?

This percentage is the pvalue of Z when $X = 83$ subtracted by the pvalue of Z when $X = 75$ . So:

X = 83

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

$Z = \frac{83 - 79}{22}$

$Z = 0.18$

$Z = 0.18$ has a pvalue of 0.5714.

X = 75

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

$Z = \frac{75- 79}{22}$

$Z = -0.18$

$Z = -0.18$ has a pvalue of 0.4286.

This means that 0.5714-0.4286 = 0.1428 = 14.28% of individual adult females have weights between 75 kg and 83 kg.

If samples of 100 adult females are randomly selected and the mean weight is computed for each sample, what percentage of the sample means are between 75 kg and 83 kg?

Now we use the Central Limit THeorem, when $n = 100$ . So $s = \frac{22}{\sqrt{100}} = 2.2.$