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3 years ago

6

Quadrilateral KLMN is rotated clockwise 90 degrees?

1 answer:

trasher [3.6K]3 years ago

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Quadrilateral KLMN is rotated clockwise 90 degrees?

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-3 (x+1) as an equivalent expression

Y_Kistochka [10]

-3x-3 is the correct answer

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3 years ago

Which of the following comparisons is true? 24,762 = 24,662 335,989 > 335,998 454,909 887,012

ratelena [41]

454,909 < 887,012 <===

8 0

3 years ago

Assume that college students have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15. Fin

jolli1 [7]

Answer:

0.018 is the probability that a randomly selected college student has an IQ greater than 131.5

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 100

Standard Deviation, σ = 15

We are given that the distribution of IQ score is a bell shaped distribution that is a normal distribution.

Formula:

a) P(IQ greater than 131.5)

P(x > 131.5)

Calculation the value from standard normal z table, we have,

0.018 is the probability that a randomly selected adult has an IQ greater than 131.5

3 0

4 years ago

NEED ANSWER ASAP PLEASE!!

erastova [34]

Answer:

DE=5*8=40

Step-by-step explanation:

DE+EF=DF

5x+17x-20=20x-4

22x-20=20x-4

22x-20x=-4+20

2x=16

x=16/2

x=8

DE=5x

DE=5*8=40

6 0

3 years ago

The following information was obtained from independent random samples taken of two populations. Assume normally distributed pop

Volgvan

Answer:

1. The 95% confidence interval for the difference between means is (-5.34, 11.34).

2. The standard error of (x-bar1)-(x-bar2) is 4.

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{9.2195^2}{10}+\dfrac{9.4868^2}{12}}\\\\\\s_{M_d}=\sqrt{8.5+7.5}=\sqrt{16}=4$

Step-by-step explanation:

We have to calculate a 95% confidence interval for the difference between means.

The sample 1, of size n1=10 has a mean of 45 and a standard deviation of √85=9.2195.

The sample 2, of size n2=12 has a mean of 42 and a standard deviation of √90=9.4868.

The difference between sample means is Md=3.

$M_d=M_1-M_2=45-42=3$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{9.2195^2}{10}+\dfrac{9.4868^2}{12}}\\\\\\s_{M_d}=\sqrt{8.5+7.5}=\sqrt{16}=4$

The critical t-value for a 95% confidence interval is t=2.086.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_{M_d}=2.086 \cdot 4=8.34$

Then, the lower and upper bounds of the confidence interval are:

$LL=M_d-t \cdot s_{M_d} = 3-8.34=-5.34\\\\UL=M_d+t \cdot s_{M_d} = 3+8.34=11.34$

The 95% confidence interval for the difference between means is (-5.34, 11.34).

6 0

4 years ago