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frozen [14]
3 years ago
5

Ruby read 0.85. Of a book what fraction of the book does ruby have left to read? Write in simplest fom

Mathematics
1 answer:
lbvjy [14]3 years ago
4 0
3/20 left to read. If you start 85/100 you take 100-85 and get 15, then you end up 15/100 which reduces to 3/20.
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What is the measure of Zx?<br> Angles are not necessarily drawn to scale.<br> B<br> 539
Ivanshal [37]
Picture of the question please?
8 0
3 years ago
How to write 2 2/13 to a decimal to nearest thousandth
Alja [10]

Answer:

2.154

Step-by-step explanation:

Convert it to an improper fraction.

Hope this helps :)

3 0
3 years ago
An article written for a magazine claims that 78% of the magazine's subscribers report eating healthily the previous day. Suppos
Schach [20]

Answer:

89.44% probability that less than 80% of the sample would report eating healthily the previous day

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.78, n = 675

So

\mu = E(X) = np = 675*0.78 = 526.5

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{675*0.78*0.22} = 10.76

What is the approximate probability that less than 80% of the sample would report eating healthily the previous day?

This is the pvalue of Z when X = 0.8*675 = 540. So

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

Z = \frac{540 - 526.5}{10.76}

Z = 1.25

Z = 1.25 has a pvalue of 0.8944

89.44% probability that less than 80% of the sample would report eating healthily the previous day

8 0
4 years ago
HELP ME PLEASE AND JUST GIVE ME THE ANSWER NO EXPLANATION
Arisa [49]

The <em>missing</em> end of the <em>line</em> segment \overline {GH} whose midpoint is M(x, y) = (5, 3) and known endpoint is G(x, y) = (1, -5) is the point H(x, y) = (9, 11).

<h3>How to locate the missing end of a line segment</h3>

In this question we know an end and a midpoint of a <em>line</em> segment. The <em>missing</em> end can be found from the following formula:

M(x, y) = G(x, y) + 0.5 · [H(x, y) - G(x, y)]

M(x, y) = 0.5 · G(x, y) + 0.5 · H(x, y)

0.5 · H(x, y) = M(x, y) - 0.5 · G(x, y)

H(x, y) = 2 · M(x, y) - G(x, y)     (1)

If we know that M(x, y) = (5, 3) and G(x, y) = (1, -5), then the coordinates of the point H are:

H(x, y) = 2 · (5, 3) - (1, -5)

H(x, y) = (10, 6) - (1, -5)

H(x, y) = (9, 11)

The <em>missing</em> end of the <em>line</em> segment \overline {GH} whose midpoint is M(x, y) = (5, 3) and known endpoint is G(x, y) = (1, -5) is the point H(x, y) = (9, 11).

To learn more on line segments: brainly.com/question/25727583

#SPJ1

3 0
2 years ago
Read 2 more answers
A x 4 = E
lana [24]
A * 4 = E........A = E / 4
B / 4 = E........B = 4E
C + 4 = E......C = E - 4
D - 4 = E.......D = E + 4

A + B + C + D = 100
(E/4) + (4E) + (E - 4) + (E + 4) = 100
E/4 + 6E = 100 ....multiply by 4
E + 24E = 400
25E = 400
E = 400/25
E = 16 <===


7 0
3 years ago
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