All categories

2 years ago

A number z squared, decreased by 7

Mathematics

1 answer:

Vedmedyk [2.9K]2 years ago

3 0

9514 1404 393

Answer:

z² -7

Step-by-step explanation:

A square is indicated using an exponent of 2. "Decreased by 7" means 7 is subtracted.

z squared, decreased by 7 is ...

z² - 7

You might be interested in

If two matrices, A and B, are equal, which of the

alexandr402 [8]

If A and B are equal:

Matrix A must be a diagonal matrix: FALSE.

We only know that A and B are equal, so they can both be non-diagonal matrices. Here's a counterexample:

$A=B=\left[\begin{array}{cc}1&2\\4&5\\7&8\end{array}\right]$

Both matrices must be square: FALSE.

We only know that A and B are equal, so they can both be non-square matrices. The previous counterexample still works

Both matrices must be the same size: TRUE

If A and B are equal, they are literally the same matrix. So, in particular, they also share the size.

For any value of i, j; aij = bij: TRUE

Assuming that there was a small typo in the question, this is also true: two matrices are equal if the correspondent entries are the same.

8 0

3 years ago

Read 2 more answers

Write an equation that expresses the following relationship.

stepladder [879]

Answer:

w = $\frac{ku}{d^{2} }$

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

1. What was the second of two crucial events that increased U.S. concerns about the spread of

sineoko [7]

1. the answer is D
2. You are correct it is vietnam
3. the answer is 1947

4 0

3 years ago

Read 2 more answers

The moon is a sphere with radius of 959 km. Determine an equation for the ellipse if the distance of the satellite from the surf

Sergeeva-Olga [200]

Answer:

$\frac{x^2}{1316^2}+\frac{y^2}{1669^2}=1$

Step-by-step explanation:

An ellipse is the locus of a point such that its distances from two fixed points, called foci, have a sum that is equal to a positive constant.

The equation of an ellipse with a center at the origin and the x axis as the minor axis is given by:

$\frac{x^2}{b^2}+\frac{y^2}{a^2} =1 \\\\where\ a>b$

Since the distance of the satellite from the surface of the moon varies from 357 km to 710 km, hence:

b = 357 km + 959 km = 1316 km

a = 710 km + 959 km = 1669 km

Therefore the equation of the ellipse is:

$\frac{x^2}{1316^2}+\frac{y^2}{1669^2}=1$

5 0

3 years ago

The mean consumption of water per household in a city was 1425 cubic feet per month. Due to a water shortage because of a drough

liberstina [14]

Answer:

a) $t=\frac{1175-1425}{\frac{250}{\sqrt{100}}}=-10$

The degrees of freedom are given by:

$df=n-1=100-1=99$

The p value for this case would be given by:

$p_v =P(t_{99}$

Since the p value is significantly lower than he significance level given we have enough evidence to reject the null hypothesis and we can conclude that the true mean is lower than 1425

b) For this case we need to find a critical value in the t distribution with 99 degrees of freedom who accumulates 0.025 of the area in the right tail and we got:

$t_{\alpha/2}=1.984$

Since the calculated value is higher than the critical value we can reject the null hypothesis at the significance level provided and we can say that the true mean is higher than 1425

Step-by-step explanation:

Information given

$\bar X=1175$ represent the sample mean for the cubic feets of households

$\sigma=250$ represent the population standard deviation

$n=100$ sample size

$\mu_o =1425$ represent the value to verify

$\alpha=0.025$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Part a

We want to test that the mean consumption of water per household has decreased due to the campaign by the city council, the system of hypothesis would be:

Null hypothesis: $\mu \geq 1425$

Alternative hypothesis: $\mu < 1425$

Since we don't know the deviation the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

The statistic is given by:

$t=\frac{1175-1425}{\frac{250}{\sqrt{100}}}=-10$

The degrees of freedom are given by:

$df=n-1=100-1=99$

The p value for this case would be given by:

$p_v =P(t_{99}$

Since the p value is significantly lower than he significance level given we have enough evidence to reject the null hypothesis and we can conclude that the true mean is lower than 1425

Part b

For this case we need to find a critical value in the t distribution with 99 degrees of freedom who accumulates 0.025 of the area in the right tail and we got:

$t_{\alpha/2}=1.984$

Since the calculated value is higher than the critical value we can reject the null hypothesis at the significance level provided and we can say that the true mean is higher than 1425

6 0

3 years ago