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BARSIC [14]
3 years ago
8

Please help and please show workwill give brainlyist

Mathematics
2 answers:
Komok [63]3 years ago
7 0

If I am understanding this right, since the <em>c </em>is a variable, you'd have to solve the equation and have the variable in the and result.

the sqrt525^7 = 3359983566.19

So you'd have to simplify this and add the <em>c</em> to the end of it. Unless your work is fractions.

(I take Algebra II I should be more confident on this topic, this is just what makes the most sense)

Alona [7]3 years ago
3 0

Answer:

5c^3\sqrt{21c}

Step-by-step explanation:

525 = 3 * 7 * 5 * 5

\sqrt{525c^7} =

= \sqrt{5^2 \times 21 \times c^6 \times c}

= 5c^3\sqrt{21c}

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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

xc mean VOT for children =60.67

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

           = √189.72559

           = 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

(c) Calculating the p-value, we have;

Degree of freedom = na + nc -2

                                 = 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.

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

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(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
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