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Harlamova29_29 [7]

2 years ago

10

Draw the image of quadrilateral ABCD under the translation (x,y) ——> (x + 4, y - 3)

1 answer:

salantis [7]2 years ago

4 0

Answer:

2 to 5

Step-by-step explanation:

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An automobile manufacturer claims that its jeep has a 53.9 miles/gallon (MPG) rating. An independent testing firm has been contr

Advocard [28]

Answer:

Value of the test statistic, $z_{test} = - 1.17$

Step-by-step explanation:

Null hypothesis, $H_{0}: \mu = 53.9$

Alternative hypothesis, $H_{a} : \mu \neq 53.9$

Sample mean, $\bar{X} = 53.7$

Sample size, n = 110

Standard deviation, $\sigma = 1.8$

Significance level, $\alpha = 0.02$

The value of the test statistics is given by the formula:

$z_{test} = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n} } } \\z_{test} = \frac{53.7 - 53.9}{\frac{1.8}{\sqrt{110} } } \\z_{test} = - 1.17$

7 0

2 years ago

Use linear approximation to approximate √25.3 as follows.

Sophie [7]

The idea is to use the tangent line to $f(x)=\sqrt x$ at $x=25$ in order to approximate $f(25.3)=\sqrt{25.3}$ .

We have

$f(x)=\sqrt x\implies f(25)=\sqrt{25}=5$

$f'(x)=\dfrac1{2\sqrt x}\implies f'(25)=\dfrac1{10}$

so the linear approximation to $f(x)$ is

$L(x)=f(5)+f'(5)(x-5)=5+\dfrac{x-5}{10}=\dfrac x{10}+\dfrac92$

Hence $m=\frac1{10}$ and $b=\frac92$ .

Then

$f(25.3)\approx L(25.3)=\dfrac{25.3}{10}+\dfrac92=\boxed{7.03}$

4 0

2 years ago

Can someone please help me:)

Naddika [18.5K]

For a translation 4 units down you would subtract 4 from each y value and 5 units right you would add 5 to each x value.
Your resulting points would be.
A'(-1,1), B'(-1,-2), C'(3,-2), D'(3,2)

4 0

2 years ago

I need help finding the domain and range

solmaris [256]

Domain is X and Range is Y

4 0

2 years ago

What is the answer to this. Please help ASAP.

Zigmanuir [339]

V of rec. prism = whl = 3x5x6 = 90
V of cone = 3.14 x (3)^2 (8) /3= 75.36

90 - 75.36 = 14.64

answer
14.64

6 0

2 years ago