What is 103 + 343 + 252

What is the common difference for the arithmetic sequence -5,-1,3,7

weeeeeb [17]

Answer: 4

-5+4=-1

-1+4=3

3+4=7

I hope this is good enough:

4 0

2 years ago

It is given that 2^x = 5^y = 10^z , express z in terms of x and y

weqwewe [10]

9514 1404 393

Answer:

z = (xy)/(x+y)

Step-by-step explanation:

Taking logarithms base 10, we have ...

x·log(2) = y·log(5) = z·log(10)

z = x·log(2) = y·log(5) . . . . . . . . log(10) = 1

__

Then z/x = log(2), and z/y = log(5), so ...

z/x +z/y = log(2) +log(5) = log(2·5) = 1

z = 1/(1/x +1/y) = 1/((x+y)/(xy))

z = (xy)/(x+y)

4 0

2 years ago

Read 2 more answers

Brainlist at the first one to answer!!!!!! Plz........ASAP!!!

Makovka662 [10]

Answer: ellipse

Step-by-step explanation:

3 0

2 years ago

Many elementary school students in a school district currently have ear infections. A random sample of children in two different

Marta_Voda [28]

Answer:

Step-by-step explanation:

The summary of the given data includes;

sample size for the first school $n_1$ = 42

sample size for the second school $n_2$ = 34

so 16 out of 42 i.e $x_1$ = 16 and 18 out of 34 i.e $x_2$ = 18 have ear infection.

the proportion of students with ear infection Is as follows:

$\hat p_1 = \dfrac{16}{42}$ = 0.38095

$\hat p_2 = \dfrac{18}{34}$ = 0.5294

Since this is a two tailed test , the null and the alternative hypothesis can be computed as :

$H_0 :p_1 -p_2 = 0 \\ \\ H_1 : p_1 - p_2 \neq 0$

level of significance ∝ = 0.05,

Using the table of standard normal distribution, the value of z that corresponds to the two-tailed probability 0.05 is 1.96. Thus, we will reject the null hypothesis if the value of the test statistics is less than -1.96 or more than 1.96.

The test statistics for the difference in proportion can be achieved by using a pooled sample proportion.

$\bar p = \dfrac{x_1 +x_2}{n_1 +n_2}$

$\bar p = \dfrac{16 +18}{42 +34}$

$\bar p = \dfrac{34}{76}$

$\bar p = 0.447368$

$\bar p + \bar q = 1 \\ \\ \bar q = 1 -\bar p \\ \\\bar q = 1 - 0.447368 \\ \\\bar q = 0.552632$

The pooled standard error can be computed by using the formula:

$S.E = \sqrt{ \dfrac{ \bar p \bar q}{ n_1} + \dfrac{\bar p \bar p}{n_2} }$

$S.E = \sqrt{ \dfrac{ 0.447368 * 0.552632}{ 42} + \dfrac{ 0.447368 * 0.447368}{34} }$

$S.E = \sqrt{ \dfrac{ 0.2472298726}{ 42} + \dfrac{ 0.2001381274}{34} }$

$S.E = \sqrt{ 0.01177284105}$

$S.E = 0.1085$

The test statistics is ;

$z = \dfrac{\hat p_1 - \hat p_2}{S.E}$

$z = \dfrac{0.38095- 0.5294}{0.1085}$

$z = \dfrac{-0.14845}{0.1085}$

z = - 1.368

Decision Rule: Since the test statistics is greater than the rejection region - 1.96 , we fail to reject the null hypothesis.

Conclusion: There is insufficient evidence to support the claim that a difference exists between the proportions of students who have ear infections at the two schools

5 0

3 years ago

If you get a sneak peak at the salary list of your job and realize you are in the lowest bracket, which is the lowest 1%, what i

hoa [83]

Answer:

$z=-2.33$

And if we solve for a we got

$a=48000 -2.33*5500=35185$

And for this case the answer would be 35185 the lowest 1% for the salary

Step-by-step explanation:

Let X the random variable that represent the salary, and for this case we can assume that the distribution for X is given by:

$X \sim N(48000,5500)$

Where $\mu=48000$ and $\sigma=5500$

And we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.99$ (a)

$P(X$ (b)

We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.01 of the area on the left and 0.99 of the area on the right it's z=-2.33. On this case P(Z<-2.33)=0.01 and P(z>-2.33)=0.99

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-2.33$

And if we solve for a we got

$a=48000 -2.33*5500=35185$

And for this case the answer would be 35185 the lowest 1% for the salary

7 0

3 years ago