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2 years ago

Logan's monthly bank statement showed the following deposits

Mathematics

1 answer:

Leno4ka [110]2 years ago

3 0

Logan's monthly bank statement showed the following deposits

and withdrawals:

$31.34, $117.42, -$52.48, -$102.33, $64.90

If Logan's balance in the account was $82.63 at the

beginning of the month, what was the account balance at

the end of the month

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Xavier has $23.50 in his savings account. He deposits $35 every week. His mother also deposits $20 into the account every time X

andrezito [222]

Part A the coefficient is 20c and 35w, the variable is c and w representing cost and week. and the constant is 23.50.

Part B He would have 503.50 because 35(12)+60=420x+60+23.50 which would be 503.50

Part C The coefficient of the cost would change because you would be adding a 5 dollar increase!!!!!! done

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2 years ago

A survey was administered to high school seniors in Anytown. According to the survey results, fewer than 0.5% of the students dr

Feliz [49]

Answer: high test-retest reliability

Step-by-step explanation: this is because the result of the survey was thesame with the previous result, despite the space of time between when the first survey was conducted and when the second survey was conducted. There was know observable difference in result and if conducted in the next 3 months again, it will give same result, this strongly indicate that the survey has high test-retest reliability.

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1 year ago

A book costs d dollars.

bazaltina [42]

Answer:

The total amount paid by Jill = $ (3 d)

Step-by-step explanation:

Here, the cost of 1 book = $ d

Now, the number of books purchased by Jill = 3

So, the total cost of 3 books = 3 x ( Cost of 1 book)

= 3 x ($d) = 3 d

So, if Jill purchased 3 books, the total amount paid by her = $ (3 d)

3 0

1 year ago

You'd like to estimate the proportion of the 12,152 full-time undergraduate students at a university who are foreign students.

Minchanka [31]

Answer:

a) The data distribution consists of ( 7 )1's (denoting a foreign student) and ( 43 )0's (denoting a student from the U.S.).

b) The population distribution consists of the x-values of the population of 12,152 full-time undergraduate students at theuniversity, ( 6 )% of which are 1's (denoting a foreign student) and ( 94 )% of which are 0's (denoting a student from the U.S.).

c) The mean is ( 0.06 )

The standard deviation is ( 0.0336 )

The sampling distribution represents the probability distribution of the ( sample ) proportion of foreign students in a random sample of ( 50 ) students. In this case, the sampling distribution is approximately normal with a mean of ( 0.06 ) and a standard deviation of ( 0.0336 )

Step-by-step explanation:

5 0

2 years ago

Suppose you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by model

Ann [662]

Answer:

(1) The degrees of freedom for unequal variance test is (14, 11).

(2) The decision rule for the 0.01 significance level is;

If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.

If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.

(3) The value of the test statistic is 0.3796.

Step-by-step explanation:

We are given that you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by models featuring Liz Claiborne's attire with those of Calvin Klein.

The following is the amount ($000) earned per month by a sample of 15 Claiborne models;

$3.5, $5.1, $5.2, $3.6, $5.0, $3.4, $5.3, $6.5, $4.8, $6.3, $5.8, $4.5, $6.3, $4.9, $4.2 .

The following is the amount ($000) earned by a sample of 12 Klein models;

$4.1, $2.5, $1.2, $3.5, $5.1, $2.3, $6.1, $1.2, $1.5, $1.3, $1.8, $2.1.

(1) As we know that for the unequal variance test, we use F-test. The degrees of freedom for the F-test is given by;

$\text{F}_(_n__1-1, n_2-1_)$

Here, $n_1$ = sample of 15 Claiborne models

$n_2$ = sample of 12 Klein models

So, the degrees of freedom = ( $n_1-1, n_2-1$ ) = (15 - 1, 12 - 1) = (14, 11)

(2) The decision rule for 0.01 significance level is given by;

If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.

If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.

(3) The test statistics that will be used here is F-test which is given by;

T.S. = $\frac{s_1^{2} }{s_2^{2} } \times \frac{\sigma_2^{2} }{\sigma_1^{2} }$ ~ $\text{F}_(_n__1-1, n_2-1_)$

where, $s_1^{2}$ = sample variance of the Claiborne models data = $\frac{\sum (X_i-\bar X)^{2} }{n_1-1}$ = 1.007

$s_2^{2}$ = sample variance of the Klein models data = $\frac{\sum (X_i-\bar X)^{2} }{n_2-1}$ = 2.653

So, the test statistics = $\frac{1.007}{2.653 } \times 1$ ~ $\text{F}_(_1_4,_1_1_)$

= 0.3796

Hence, the value of the test statistic is 0.3796.

3 0

1 year ago