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3 years ago

Which values of

Mathematics

1 answer:

I am Lyosha [343]3 years ago

7 0

83x + P = 83x + Q subtract 83x from both sides

P = Q

The equation has no solution for P different Q (P ≠ Q).

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Annie estimates that the height of a bookcase is 78.25 in. The actual height is 75.50 in. To the nearest tenth of a percent, wha

Vinil7 [7]

Answer

Find out the percent error in Annie's estimate .

To prove

Formula

$Percentage\ error = \frac{error\times 100}{exact\ value}$

As error = approx value - exact value

As given

Annie estimates that the height of a bookcase is 78.25 in.

The actual height is 75.50 in.

approx value = 78.25 in.

actual value = 75.50 in.

error = 78 .25 - 75 .50

= 2.75

put in the formula

$Percentage\ error = \frac{2.75\times 100}{75.50}$

$Percentage\ error = \frac{27500}{7550}$

Percentage error = 3.64 % (approx)

Therefore the Percentage error be 3.64 % (approx).

6 0

3 years ago

Read 2 more answers

What is the parameter of interest to compare the proportions from two populations?

Marysya12 [62]

As with comparing two population proportions, when we compare two population means from independent populations, the interest is in the difference of the two means. In other words, if is the population mean from population 1 and is the population mean from population 2, then the difference is μ 1 − μ 2 .

Mark brainliest please

Hope this helps you

7 0

2 years ago

In the equation 23×406= 9,338. which number is the product?

krek1111 [17]

Answer:

9,338 is the product of the equation

8 0

3 years ago

Several programs attempt to address the shortage of qualified teachers by placing uncertified instructors in schools with acute

ss7ja [257]

Answer:

We conclude that the mean scores with uncertified teachers is higher or equal as compared to certified teachers.

Step-by-step explanation:

We are given that reading scores of the students of certified teachers averaged 35.62 points with standard deviation 9.31. The scores of students instructed by uncertified teachers had mean 32.48 points with standard deviation 9.43 points on the same test.

There were 44 students in each group.

Let $\mu_1$ = mean scores with uncertified teachers.

$\mu_2$ = mean scores with certified teachers.

So, Null Hypothesis, $H_0$ : $\mu_1\geq \mu_2$ {means that the mean scores with uncertified teachers is higher or equal as compared to certified teachers}

Alternate Hypothesis, $H_A$ : $\mu_1$ {means that the mean scores with uncertified teachers is lower as compared to certified teachers}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t__n__1-_n__2-2$

where, $\bar X_1$ = sample mean scores of students instructed by uncertified teachers = 32.48 points

$\bar X_2$ = sample mean scores of students instructed by certified teachers = 35.62 points

$s_1$ = sample standard deviation of scores of students instructed by uncertified teachers = 9.43 points

$s_2$ = sample standard deviation of scores of students instructed by certified teachers = 9.31 points

$n_1$ = sample of students under uncertified teachers = 44

$n_2$ = sample of students under certified teachers = 44

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(44-1)\times 9.43^{2} +(44-1)\times 9.31^{2} }{44+44-2} }$ = 9.37

So, the test statistics = $\frac{(32.48-35.62)-(0)}{9.37 \times \sqrt{\frac{1}{44} +\frac{1}{44} } }$ ~ $t_8_6$

= -1.572

The value of t test statistics is -1.572.

Since, in the question we are not given the level of significance so we assume it to be 5%. Now, at 5% significance level the t table gives critical values of -1.665 at 86 degree of freedom for left-tailed test.

Since our test statistic is more than the critical values of t as -1.572 > -1.665, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the mean scores with uncertified teachers is higher or equal as compared to certified teachers.

7 0

3 years ago

if you brought a stock last year for a price of $106,and it has gone down 15%since then,how much is the stock worth now,to the n

rusak2 [61]

Answer:

the answer is 90 dollars and 1 cent

8 0

3 years ago