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nexus9112 [7]
3 years ago
8

Please help. im not good with probability

Mathematics
1 answer:
qaws [65]3 years ago
8 0
2 out of 10 chances

2/10 is 1/5
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Ulleksa [173]

Answer:

dont use this

Step-by-step explanation:

oinf+euiv=4

7 0
3 years ago
Calculate 177537+87439-19673​
nadezda [96]

Answer:

See below.

Step-by-step explanation:

Following order of operations:

177537+87439=264976

264976-19673=245303

So, 177537+87439-19673​ is equal to 245303.

-hope it helps

8 0
2 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

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.

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.

7 0
3 years ago
What can I do to solve the problem
vitfil [10]

Multiply all the numbers together to get your answer.

7 0
3 years ago
Read 2 more answers
Octagon ABCDEFGH and its dilation, octagon A'B'C'D'E'F'G'H', are shown on the coordinate plane below:
choli [55]
We can take out one of the points such as H (2,6) and then take it's dilation H' (1,3)and see that 2 and 6 were both multiplied by 1/2 to get 1 and 3, so the scale factor is 1/2
3 0
3 years ago
Read 2 more answers
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