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3 years ago

What are all of the values?

Mathematics

1 answer:

Arada [10]3 years ago

4 0

(i) x = 3

(ii) y = 6

<u>Explanation:</u>

The two polygons are similar.

The congruent angles are:

∠J ≅ ∠G

∠R ≅ ∠M

∠I ≅ ∠A

∠C ≅ ∠T

∠E ≅ ∠H

Proportional sides are:

$\frac{JR}{GM} = \frac{RI}{MA} = \frac{IC}{AT} = \frac{CE}{TH} = \frac{EJ}{HG}$

(i) Value of x = ?

$\frac{JR}{GM} = \frac{RI}{MA}$

Putting the values from the figure

$\frac{6}{12} = \frac{2x + 1}{3x + 5} \\\\6 ( 3x + 5) = 12 ( 2x + 1)\\\\18x + 30 = 24x + 12\\\\30 - 12 = 24x - 18x\\\\18 = 6x\\\\x = 3$

(ii) Value of y = ?

$\frac{JR}{GM} = \frac{IC}{AT}$

On putting the value we get:

$\frac{6}{12} = \frac{x^2-4}{y+4} \\\\6(y+4) = 12(x^2 - 4)\\\\6y + 24 = 12[(3)^2 - 4]\\\\6y + 24 = 12[ 9 - 4]\\\\6y + 24 = 60\\\\6y = 36\\\\y = 6$

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Step-by-step explanation:

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Then the extremes are:

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Then this segment can be written as:

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Since it is given that AB ≅ AC, it must also be true that AB = AC. Assume ∠B and ∠C are not congruent. Then the measure of one a

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A plant is already 52 centimeters tall, and it will grow one centimeter every month. Let H be the plant's health (in centimeters

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<h3>How to find the height of the plant</h3>

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insurance company checks police records on 559 accidents selected at random and notes that teenagers were at the wheel in 91 of

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Critical value : $z_{\alpha/2}=1.96$

Confidence level for population proportion:-

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3 years ago

Researchers studying the acquisition of pronunciation often compare measurements made on the recorded speech of adults and child

dolphi86 [110]

Answer:

(A) The additional information that is needed to confirm about the conditions for this test have been met is ‘Population is approximately normal.

(B) The test statistic = 1.9965

(C1) P-value = 0.0556

(C2) There is no significant difference between mean VOT for children and adults.

D1) It is possible to make type II error.

(D2) There should be a difference in the VOT of adults and children.

Step-by-step explanation:

Let na be the number of adults = 20

xa mean VOT for adults = 88.17

sa standard deviatiation of VOT for adults = 24.74

Let nc be the number of children = 10

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sc standard deviatiation of VOT for children = 39.89

(A) From the information the population variances are unknown and the two sample are assumed to be independent and the sample the sample size are smaller that is (n<30).

This indicates that the additional information that is required for the conditions of the test to be satisfied is 'distribution of the population'. the addition assumption to be made is, that the population distribution is normal.

(b) Calculating the test statistics using the formula;

t = (xa -xc)/SE - d

where SE = standard deviation , d= hypothesized difference = 0

But SE = √sa²/na +sc²/nc

= √24.74 ²/20 + 39.89²/10

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= 13.774

Substituting into test statistics equation, we have

t = (xa -xc)/SE - d

= (88.17 - 60.67/13.774

= 1.9965

Therefore the test statistic is 1.9965

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= 10+ 20 -2

= 28

The p-value for t=1.9965 at 28 degrees of freedom and 0.05 level of significance is 0.0556.

The p-value 0.0556 is greater than given level of significance 0.05 hence we fail to reject the null hypothesis and conclude that there is no significant difference between mean VOT for children and adults.

(D1) From the information in part (C) the null hypothesis is not rejected.

Since the null hypothesis is not rejected, there might be a chance that not rejecting null hypothesis would be wrong. In this type of situation the error that can occur would be type II error.

(D2) The type of error describe in the context of this study is obtained by the concept of the type II error which tells that the null hypothesis is not rejected when it is actually false.

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3 years ago