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2 years ago

Use a distributive property to remove the parentheses -9(2v-u-3)

Mathematics

2 answers:

jeka57 [31]2 years ago

4 0

-9(2v-u-3)
-18v+9u+27

PIT_PIT [208]2 years ago

4 0

the answer is -18v+9u+27

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Rational formulas and variations

Zanzabum

Answer:

Direct to running

Joint to apples

Direct to meat

I hope this is good enough:

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2 years ago

How much oil wells in a given field will ultimately produce is key information in deciding whether to drill more wells. Data was

Ugo [173]

Answer:

Check the explanation

Step-by-step explanation:

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3 years ago

What is the equation in point-slope form of a line that passes through the points (7, −8) and (−4, 6) ?

Gala2k [10]

(7,-8)(-4,6)
slope = 6 - (-8) / (-4 - 7) = (6 + 8) / -11 = -14/11

point slope : y - y1= m(x - x1)
(-4,6)...x1 = -4 and y1 = 6
now we sub
y - 6 = -14/11(x - (-4) =
y - 6 = -14/11(x + 4) <===

5 0

2 years ago

Read 2 more answers

4. A horizontal ellipse, centered at the origin has a major axis of 10 units and minor axis of 8 units. Write the

s2008m [1.1K]

Answer:

$\frac{x^2}{25} +\frac{y^2}{16} =1$

Step-by-step explanation:

For ellipses, the length of the major axis is represents as:

Major axis = $2a$

where $a$ is called the semi-major axis.

In this case since the major axis is equal to 10 units:

$10=2a$

solving for the semi-major axis $a$ :

$a=10/2\\a=5$

and also the minor axis of an ellipse is represented as:

Minor axis = $2b$

where $b$ is called the semi-minor axis.

Since the minor axis has a length of 8 units:

$8=2b$

solving for b:

$b=8/2\\b=4$

Now we can use the equation for an ellipse centered at the origin (0,0):

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

and substituting the values for $a$ and $b$ :

$\frac{x^2}{5^2} +\frac{y^2}{4^2} =1$

and finall we simplify the expression to get the equation of the ellipse:

$\frac{x^2}{25} +\frac{y^2}{16} =1$

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2 years ago

A healthcare provider monitors the number of CAT scans performed each month in each of its clinics. The most recent year of data

Rasek [7]

Answer: a. 1.981 < μ < 2.18

b. Yes.

Step-by-step explanation:

A. For this sample, we will use t-distribution because we're estimating the standard deviation, i.e., we are calculating the standard deviation, and the sample is small, n = 12.

First, we calculate mean of the sample:

$\overline{x}=\frac{\Sigma x}{n}$

$\overline{x}=\frac{2.31=2.09+...+1.97+2.02}{12}$

$\overline{x}=$ 2.08

Now, we estimate standard deviation:

$s=\sqrt{\frac{\Sigma (x-\overline{x})^{2}}{n-1} }$

$s=\sqrt{\frac{(2.31-2.08)^{2}+...+(2.02-2.08)^{2}}{11} }$

s = 0.1564

For t-score, we need to determine degree of freedom and $\frac{\alpha}{2}$ :

df = 12 - 1

df = 11

$\alpha$ = 1 - 0.95

α = 0.05

$\frac{\alpha}{2}=$ 0.025

Then, t-score is

$t_{11,0.025}$ = 2.201

The interval will be

$\overline{x}$ ± $t.\frac{s}{\sqrt{n} }$

2.08 ± $2.201\frac{0.1564}{\sqrt{12} }$

2.08 ± 0.099

The 95% two-sided CI on the mean is 1.981 < μ < 2.18.

B. We are 95% confident that the true population mean for this clinic is between 1.981 and 2.18. Since the mean number performed by all clinics has been 1.95, and this mean is less than the interval, there is evidence that this particular clinic performs more scans than the overall system average.

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3 years ago