(-327) + 420 + 521=

Solve for x. This figure is a trapezoid. A) 12 B)9 C)3 D) 10

kicyunya [14]

Answer:

c)3

Step-by-step explanation:

numbers between 6 and 10 are, 7, 8, and 9

the median is 8 so the line in the middle has to be 8

8-5=3 so x=3

7 0

3 years ago

Read 2 more answers

Simplify the expression 9xy + 6xy − 2xy. What is the coefficient of the simplified expression?

Art [367]

Hi there!

To find our answer we need to collect the terms of the expression.

$9xy + 6xy - 2xy = \\ 15xy - 2xy = \\ 13xy$

The coefficient (which is the number just before the variable) of the simplified expression is 13.

3 0

4 years ago

Read 2 more answers

When a dividend greater than 1 is divided by a divisor less than 1, the quotient is

marysya [2.9K]

Answer:

Dividing by 1 gives a quotient equal to the dividend. When the divisor is less than 1, the quotient is larger than the dividend. Decreasing the divisor to 1/2 increases the quotient to 10 1/2 When the divisor is smaller than the dividend, the quotient is more than 1.

7 0

3 years ago

Turn 258/300 into a decimal and percent

erastova [34]

Simply divide the numerator by the denominator.

258 / 300 = 0.86 (as a decimal)

Now, multiply the decimal by 100 to find the percent.

0.86 * 100 = 86%

Hope this helps!

3 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