Help. Me with this please.

Nia increased her class average by 3% she now has 92% in her math class

hichkok12 [17]

Answer:

see below

Step-by-step explanation:

1.03 * x = 92 x = prior percentage

92 / 1.03 = x = 89.32

She HAD a 89.32 percentage score

increasing this score 3% reults in a score of 92

6 0

1 year ago

Find the mass of the solid paraboloid Dequals={(r,thetaθ,z): 0less than or equals≤zless than or equals≤8181minus−r2, 0less th

Lubov Fominskaja [6]

Answer:

M = 5742π

Step-by-step explanation:

Given:-

- Find the mass of a solid with the density ( ρ ):

ρ ( r, θ , z ) = 1 + z / 81

- The solid is bounded by the planes:

0 ≤ z ≤ 81 - r^2

0 ≤ r ≤ 9

Find:-

Find the mass of the solid paraboloid

Solution:-

- The mass (M) of any solid body is given by the following triple integral formulation:

$M = \int \int \int {p ( r ,theta, z)} \, dV\\\\$

- We can write the above expression in cylindrical coordinates:

$M = \int\limits\int\limits_r\int\limits_z {r*p(r,theta,z)} \, dz.dr.dtheta \\\\M = \int\limits\int\limits_r\int\limits_z {r*[ 1 + \frac{z}{81}] } \, dz.dr.dtheta\\\\$

- Perform integration:

$M = \int\limits\int\limits_r{r*[ z + \frac{z^2}{162}] } \,|_0^8^1^-^r^2 dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + \frac{(81-r^2)^2}{162}] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + \frac{6561 -162r + r^2}{162}] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + 40.5 -r +\frac{r^2}{162} ] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{[ 121.5r-r^2 -\frac{161r^3}{162} ] } \, dr.dtheta\\\\$

$M = 2*\int\limits_0^\pi {[ 121.5r^2-r^3 -\frac{161r^4}{162} ] } |_0^6 \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 121.5(6)^2-(6)^3 -\frac{161(6)^4}{162} ] } \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 4375-216 -1288] } \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 2871] } \, dtheta\\\\M = 5742\pi kg$

- The mass evaluated is M = 5742π

8 0

2 years ago

A survey in Men’s Health magazine reported that 39% of cardiologists said that they took vitamin E supplements. To see if this i

zzz [600]

Answer:

$z=\frac{0.36 -0.39}{\sqrt{\frac{0.39(1-0.39)}{100}}}=-0.615$

The p value for this case would be:

$p_v =2*P(z$

For this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true proportion is not different from 0.39

Step-by-step explanation:

Information given

n=100 represent the random sample taken

X=36 represent the number of people that take E supplement

$\hat p=\frac{36}{100}=0.36$ estimated proportion of people who take R supplement

$p_o=0.39$ is the value that we want to test

$\alpha=0.05$ represent the significance level

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true proportion is equatl to 0.39 or not, the system of hypothesis are.:

Null hypothesis: $p=0.39$

Alternative hypothesis: $p \neq 0.39$

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing the info we got:

$z=\frac{0.36 -0.39}{\sqrt{\frac{0.39(1-0.39)}{100}}}=-0.615$

The p value for this case would be:

$p_v =2*P(z$

For this case since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true proportion is not different from 0.39

8 0

2 years ago

Can someone help with geometry study

Studentka2010 [4]

This question is incomplete.

6 0

3 years ago

Read 2 more answers

In the rhombus,8y-6 . Find the value of y.

gregori [183]

I think that the answer to the question is 0.75

7 0

2 years ago