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3 years ago

I need help with all 3 please

Mathematics

1 answer:

olganol [36]3 years ago

6 0

Step-by-step explanation:

x +2y = 9. .....(1)

x - 2y = 5. .....(2)

from eqn (1)

x + 2y = 9

x = 9 - 2y.

substitute x = 9 - 2y into eqn (2)

9 - 2y - 2y = 5

-4y = 5 - 9

-4y = -4

y = 1

sub y = 1 into eqn (1)

x + 2y = 9

x + 2(1) = 9

x + 2 = 9

x = 9 - 2

x = 7.

(2) x + y = 9 ....(1)

x - 2y = 0 .....(2)

from eqn (2)

x - 2y = 0

x =0 + 2y

substitute x = 2y into eqn (1)

x + y = 9

(2y) + y = 9

3y = 9

y = 3

substitute y = 3 into eqn (2)

x - 2y = 0

x - 2(3) = 0

x - 6 = 0

x = 6.

(3) 2x + 7y = 5. ....(1)

2x + 3y = 9. .....(2)

from eqn (1)

2x + 7y = 5

7y = 5 - 2x

y = (5 - 2x)/7

sub y = (5 - 2x)/7 into eqn (2)

2x + 3y = 9

2x + 3(5 - 2x)/7 = 9

2x + (15 - 6x)/7 = 9

multiply through by 7

14x + 15 - 6x = 63

14x - 6x = 63 - 15

8x = 48

x = 6

sub x = 6 into eqn (1)

2x + 7y = 5

2(6) + 7y = 5

12 + 7y = 5

7y = 5 - 12

7y = -7

y = -1

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Time spent using e-mail per session is normally distributed, with mu equals 11 minutes and sigma equals 3 minutes. Assume that

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Answer:

a) 0.259

b) 0.297

c) 0.497

Step-by-step explanation:

To solve this problem, it is important to know the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central limit theorem

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 $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 11, \sigma = 3$

a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that $n = 25, s = \frac{3}{\sqrt{25}} = 0.6$

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

$Z = \frac{11.2 - 11}{0.6}$

$Z = 0.33$

$Z = 0.33$ has a pvalue of 0.6293.

X = 10.8

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

$Z = \frac{10.8 - 11}{0.6}$

$Z = -0.33$

$Z = -0.33$ has a pvalue of 0.3707.

0.6293 - 0.3707 = 0.2586

0.259 probability, rounded to three decimal places.

b. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.5 and 11 minutes?

Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So

X = 11

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

$Z = \frac{11 - 11}{0.6}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

X = 10.5

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

$Z = \frac{10.5 - 11}{0.6}$

$Z = -0.83$

$Z = -0.83$ has a pvalue of 0.2033.

0.5 - 0.2033 = 0.2967

0.297, rounded to three decimal places.

c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that $n = 100, s = \frac{3}{\sqrt{100}} = 0.3$

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

$Z = \frac{11.2 - 11}{0.3}$

$Z = 0.67$

$Z = 0.67$ has a pvalue of 0.7486.

X = 10.8

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

$Z = \frac{10.8 - 11}{0.3}$

$Z = -0.67$

$Z = -0.67$ has a pvalue of 0.2514.

0.7486 - 0.2514 = 0.4972

0.497, rounded to three decimal places.

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If a girl read 3/8 of a chapter book and another girl read 3/6 of a chapter book which girl read more pages

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Answer:

The girl which read 3/6 of a chapter book read more pages than the other girl.

Step-by-step explanation:

We need to compare between $\frac{3}{8} \ and \ \frac{3}{6}$

so, the denominator of both fractions should be the same number

Factor of 6 = 1 * 2 * * 3

Factor of 8 = 1 * 2 * 2 * 2

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The common denominator = 1 * 2 * 2 * 2 * 3 = 24

$\frac{3}{8} = \frac{3*3}{3*8} =\frac{9}{24} \\\frac{3}{6} = \frac{4*3}{4*6} =\frac{12}{24}$

Then compare between $\frac{9}{24} \ and \frac{12}{24} \\$

So, the fraction of the great numerator will be the great fraction

$\frac{12}{24} > \frac{9}{24}$

∴ $\frac{3}{6} > \frac{3}{8}$

==================================================

<u>Another solution:</u>

Convert both fractions to decimal:

3/8 = 0.375

3/6 = 0.5

0.5 > 0.375

===================================================

<u>Third Solution: </u>

As the numerators are equal

The fraction of least denominator is the great fraction

So, 3/6 > 3/8

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3 years ago