Keusisdnrnsusjdhhrnrksysskrkridbebrb thank you

Help if your smart plzzzzzzz <br> ( will give 35 pts and brainliest)

Temka [501]

Answer:

a) alternate interior angles theorem

b) OXP ≅ XOL

c) XO ≅ OX

d) reflexive property (i'm not sure about this one)

e) ΔXOP ≅ ΔOXL

f) cpctc

make sure to double check the fourth one

4 0

2 years ago

Will mark brainlyist or however you spell it

eimsori [14]

Your answer should be C! Hope that helps

6 0

3 years ago

If log a = x and log b = y, then log (ab2) equals

mylen [45]

Answer:

The answer for log(ab²) is x + 2y.

Step-by-step explanation:

You have to apply Logarithm Law,

$log(a) + log(b) \: ⇒ \: log(a \times b)$

$log( {a}^{n} ) \: ⇒ \: n log(a)$

In this question, you have to seperate it out :

$log(a {b}^{2} ) = log(a) + log( {b}^{2} )$

$log( {b}^{2} ) = 2 log(b)$

$let \: log(a) = x$

$let log(b) = y$

$log(a {b}^{2} ) = log(a) + 2 log(b)$

$log(a {b}^{2} ) = x + 2y$

3 0

3 years ago

A satellite is composed of 30 modular units, each of which is equipped with a set of sensors, some of which have been upgraded.

Mazyrski [523]

Answer:

1/7 (option d) of the sensors on the satellite have been upgraded

Step-by-step explanation:

Each unit contains the same number of non-upgraded sensors

number of non-upgraded sensors for each module (nus)

total number of upgraded sensors on the satellite (tus)

satellite is composed of 30 modular units

total number of non-upgraded sensors on the satellite (tnus):

tnus=30*nus

total number of sensors on the satellite (ts):

ts=tnus+tus = 30*nus + tus (I)

The number of non-upgraded sensors on one unit is 1/5 the total number of upgraded sensors on the entire satellite

nus=(1/5)*tus

tus = 5 * nus (II)

Fraction of the sensors on the satellite have been upgraded (FU):

FU = tus/ts

Using I and II

FU= (5* nus)/(30*nus + tus)

FU = (5* nus)/(30*nus + 5 * nus)

FU = (5* nus)/(35*nus)

FU = 1/7

1/7 (option d) of the sensors on the satellite have been upgraded

6 0

3 years ago

The standard deviation of math test scores at one high school is 16.1. A teacher claims that the standard deviation of the girls

Elanso [62]

Answer:

$\chi^2 =\frac{22-1}{259.21} 166.41 =13.482$

$p_v =P(\chi^2$

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(13.482,21,TRUE)"

If we compare the p value and the significance level provided we see that $p_v >\alpha$ so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population deviation is not significantly lower than 16.1 at 1% of significance.

Step-by-step explanation:

Notation and previous concepts

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

$n=22$ represent the sample size

$\alpha=0.01$ represent the confidence level

$s^2 =12.9^2=166.41$ represent the sample variance obtained

$\sigma^2_0 =16.1^2 =259.21$ represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population deviation is smaller than 16.1 (that's equivalent to check if the population variance is lower than 259.21:

Null Hypothesis: $\sigma^2 \geq 259.21$

Alternative hypothesis: $\sigma^2$

Calculate the statistic

For this test we can use the following statistic:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

$\chi^2 =\frac{22-1}{259.21} 166.41 =13.482$

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 21. And since is a left tailed test the p value would be given by:

$p_v =P(\chi^2$

In order to find the p value we can use the following code in excel:

"=CHISQ.DIST(13.482,21,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that $p_v >\alpha$ so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population deviation is not significantly lower than 16.1 at 1% of significance.

3 0

3 years ago

Keusisdnrnsusjdhhrnrksysskrkridbebrb thank you​

Keusisdnrnsusjdhhrnrksysskrkridbebrb thank you