The answer and I will mark Brainalist

What are the coordinates of the midpoint between (3, -5) and (-7, 2)?

Dahasolnce [82]

Answer:

(-2,-3/2)

Step-by-step explanation:

Midpoint formula=(x1+x2/2, y1+y2/2)

So...

(3+-7/2, -5+2/2) =

(-2, -3/2)

5 0

3 years ago

Read 2 more answers

The line plot shows how many siblings each student has. how many more students have only one sibling than at least 3 siblings

Vaselesa [24]

So for this you must ignore the student with 2.

then you see how many in order

1-6 siblings
1-5 siblings
2-3 siblings

add up the left coulumn to get 4 people have more than 3 siblings

8 0

4 years ago

$67.50 with 8% sales tax

inna [77]

Answer:

The answer is 73.05

Step-by-step explanation:

To find the answer simply multiply 67.50 by 0.08 and then add that to 67.50

and since we are talking about dollars round to the nearest cent

6 0

3 years ago

Rothamsted Experimental Station (England) has studied wheat production since 1852. Each year many small plots of equal size but

podryga [215]

Answer:

(a) 0.653

(b) 0.0198

(c) Yes, after testing we conclude that the population variance of annual wheat production for the first plot is larger than that for the second plot.

Step-by-step explanation:

(a) We are give the sample of one year annual production of wheat (in pounds) ;

4.46, 4.21, 4.40, 4.81, 2.81, 2.90, 4.93, 3.54, 4.16, 4.48, 3.26, 4.74, 4.97, 4.02, 4.91, 2.59

For calculating sample variance, firstly we will calculate mean of the above data;

Mean of above data, $X_1bar$ = Sum of all values ÷ n (no. of values)

= $\frac{4.46 + 4.21+4.40+.......+4.91+2.59}{16}$ = 4.074

Sample Variance, $s_1^{2}$ = $\frac{\sum (X-X_1bar)^{2} }{n-1}$ = $\frac{(4.46-4.074)^{2}+(4.21-4.074)^{2}+.........+(4.91-4.074)^{2}+(2.59-4.074)^{2} }{16-1}$ = 0.653

(b) Another sample for annual wheat production (in pounds);

3.89, 3.81, 3.95, 4.07, 4.01, 3.73, 4.02, 3.78, 3.72, 3.96, 3.62, 3.76, 4.02, 3.73, 3.94, 4.03

Mean of above data, $X_2bar$ = Sum of all values ÷ n (no. of values)

= $\frac{3.89+3.81+3.95.......+3.94+4.03}{16}$ = 3.88

Sample Variance, $s_2^{2}$ = $\frac{\sum (X-X_2bar)^{2} }{n-1}$ = $\frac{(3.89-3.88)^{2}+(3.81-3.88)^{2}+.........+(3.94-3.88)^{2}+(4.03-3.88)^{2} }{16-1}$ = 0.0198

(c) Now, we have to test the claim that population variance of annual wheat production for the first plot is larger than that for the second plot i.e.;

Null Hypothesis, $H_0$ : $\sigma_1^{2} = \sigma_2^{2}$ or $H_0$ : $\frac{\sigma_1^{2} }{\sigma_2^{2} } = 1$

Alternate Hypothesis, $H_0$ : $\sigma_1^{2} > \sigma_2^{2}$ or $H_0$ : $\frac{\sigma_1^{2} }{\sigma_2^{2} } > 1$

The test statistics used here is;

$\frac{s_1^{2} }{s_2^{2} }* \frac{\sigma_1^{2} }{\sigma_2^{2}}$ ~ $F_n__1-1_,n_2-1$ where, $n_1 = 16$ and $n_2 =16$

Test Statistics = $\frac{0.653 }{ 0.0198}* 1$ ~ $F_1_5_,_1_5$

= 32.98

Since, we are not provided with any significance level so we assume it to be 5% and at this level, the F table gives critical value of 2.4282.

Since our test statistics is higher than the critical value and it falls in the rejection region so we have sufficient evidence to reject null hypothesis and conclude that the population variance of annual wheat production for the first plot is larger than that for the second plot.

8 0

3 years ago

What is the value of the sum? - 3/8+ 5/10 -11/16 -1/16 1/16 1/4 -1/4

mafiozo [28]

First, you must convert the decimals to have the same denominator.

Find the least common multiple of both fractions' denominators:

Multiples of 8: {8,16,24,32,40}
Multiples of 10: {10,20,30,40}

The least common multiple of 8 and 10 is 40.

Multiply the fractions by a number that will make their denominator 40:

$- \frac{3}{8} \times \frac{5}{5} = - \frac{15}{40}$
$\frac{5}{10} \times \frac{4}{4} = \frac{20}{40}$

Now you can add these fractions together:

$-\frac{15}{40} + \frac{20}{40} = \frac{5}{40}$

Simplify this fraction:

$\frac{5}{40} \div \frac{5}{5} = \frac{1}{8}$

The answer should be 1/8.

7 0

3 years ago