Which inequality is graphed below

−8.5x + 0.36 = −3.04

erma4kov [3.2K]

X= 0.4

-8.5x=-3.4
x=0.4

4 0

2 years ago

Can someone help me to find length CD

uysha [10]

Answer:

CD = 3.602019190339

Step-by-step explanation:

CD = DA - CA

DA = DB×Cos(29) = 18.7×cos(29) = 16.355388523507

BA = BA×cos(43) = 18.7×cos(43) = 13.676314220278

CA = BA÷tan(47) = 13.676314220278÷tan(47) = 12.753369333168

Then

CD = 16.355388523507 - 12.753369333168 = 3.602019190339

7 0

3 years ago

Multiply (4.7 x 104) by (6.4 x 1015). Express your answer in scientific notation. A) 3.008 x 1020 B) 3.008 x 1060 C) 3.008 x 106

liq [111]

It's answer A, 3.008 x 10^20

6 0

3 years ago

5x=1/2x does it equal 0 ?

lorasvet [3.4K]

Answer:

Yes it does equal zero.

4 0

3 years ago

A psychologist is interested in knowing whether adults who were bullied as children differ from the general population in terms

MissTica

Answer:

$t = \frac{39.5-30.6}{\frac{6.6}{\sqrt{25}}}= 6.74$

The degrees of freedom are given by:

$df =n-1= 25-1=24$

Now we can calculate the p value with the following probability:

$p_v = 2*P(t_{24}>6.74)= 5.69x10^{-7}$

And for this case since the p value is lower compared to the significance level $\alpha=0.05$ we can reject the null hypothesis and we can conclude that the true mean for this case is different from 30.6 at the significance level of 0.05

Step-by-step explanation:

For this case we have the following info given:

$\bar X = 39.5$ represent the sample mean

$s =6.6$ represent the sample deviation

$\mu = 30.6$ represent the reference value to test.

$n = 25$ represent the sample size selected

The statistic for this case is given by:

$t =\frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}$

And replacing we got:

$t = \frac{39.5-30.6}{\frac{6.6}{\sqrt{25}}}= 6.74$

The degrees of freedom are given by:

$df =n-1= 25-1=24$

Now we can calculate the p value with the following probability:

$p_v = 2*P(t_{24}>6.74)= 5.69x10^{-7}$

And for this case since the p value is lower compared to the significance level $\alpha=0.05$ we can reject the null hypothesis and we can conclude that the true mean for this case is different from 30.6 at the significance level of 0.05

3 0

3 years ago