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CaHeK987 [17]
3 years ago
11

Please help and thank you

Mathematics
2 answers:
tatiyna3 years ago
5 0

Answer:

b_{1}=\frac{2A}{h}-b_{2}

Step-by-step explanation:

Given: A=\frac{1}{2}(b_{1} +b_{2})h

We need to completely isolate b_{1} to solve.

A=\frac{1}{2}(b_{1} +b_{2})h

A=(\frac{1}{2}b_{1} +\frac{1}{2} b_{2})h

A=\frac{1}{2}b_{1} h+\frac{1}{2}b_{2}h

-\frac{1}{2}b_{1}h+A=\frac{1}{2}b_{2}h

-\frac{1}{2}b_{1}h=-A+\frac{1}{2}b_{2}h

-\frac{1}{2}b_{1}=\frac{-A}{h}+\frac{1}{2}b_{2}

Finally, multiply both sides by -2 to completely isolate b_{1}.

b_{1}=\frac{2A}{h}-b_{2}

Marysya12 [62]3 years ago
4 0
<h2>Answer:</h2>

<u>The right option is </u><u>D</u>

<h2>Step-by-step explanation:</h2>

See the image

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Let μ denote the true average radioactivity level(picocuries per liter). The value 5 pCi/L is considered thedividing line betwee
VMariaS [17]

Answer:

H0: μ = 5 versus Ha: μ < 5.

Step-by-step explanation:

Given:

μ = true average radioactivity level(picocuries per liter)

5 pCi/L = dividing line between safe and unsafe water

The recommended test here is to test the null hypothesis, H0: μ = 5 against the alternative hypothesis Ha: μ < 5.

A type I error, is an error where the null hypothesis, H0 is rejected when it is true.

We know type I error can be controlled, so safer option which is to test H0: μ = 5 vs Ha: μ < 5 is recommended.

Here, a type I error involves declaring the water is safe when it is not safe. A test which ensures that this error is highly unlikely is desirable because this is a very serious error. We prefer that the most serious error be a type I error because it can be explicitly controlled.

7 0
3 years ago
Write the word sentence as an equation. Then solve.
Tresset [83]

Answer:

yes

Step-by-step explanation:

4 0
3 years ago
T-activity-player/star/start
wolverine [178]

Answer:

$2400

Step-by-step explanation:

as you realize, it is going down by $400 per year

6000, 5600, 5200, 4800, 4400, 4000, 3600, 3200, 2800, 2400
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3 0
2 years ago
Suppose that the function f is defined. , for all real numbers, as follows.
aivan3 [116]

Answer:   <em>The values of......</em>

<em>f(-5)=\frac{9}{4}\\ \\ f(-2)= -1\\ \\ f(4)=0</em>

Step-by-step explanation:

Given function is:   \left \{ {{f(x)= \frac{1}{4}x^2 -4, if x\neq -2} \atop {=-1, if x=-2}} \right.

For f(-5),  the value of x is -5, which is not equal to -2.

So,  f(-5)= \frac{1}{4}(-5)^2 -4 = \frac{1}{4}(25)-4 =\frac{25}{4}-4=\frac{25-16}{4}=\frac{9}{4}

For f(-2),  the value of x is -2.

So,  f(-2)= -1

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So,  f(4)=\frac{1}{4}(4)^2 -4 =\frac{1}{4}(16)-4 =4-4=0

5 0
4 years ago
9
katrin2010 [14]

Answer:

230 - 151 + 180 + (43 - 12) = 290

Step-by-step explanation:

Use PEMDAS.

Evaluate the expression in the parentheses:

230 - 151 + 180 + (43 - 12)

43 - 12 = 31

230 - 151 + 180 + 31

Add and Subtract From Left to Right:

230 - 151 + 180 + 31

79 + 180 + 31

259 + 31

290

<em>None of the given options are correct. </em>

6 0
3 years ago
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