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PtichkaEL [24]
3 years ago
10

How to do it I don’t know how to do it

Mathematics
1 answer:
MaRussiya [10]3 years ago
4 0

Answer:

Substitute

Step-by-step explanation:

(3,1) The three goes anywhere theres an x

The 1 goes anywhere theres a y


Hope this helps

May I please be brainliest?


~Courtney

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Evaluate this exponential expression.<br> 4. (2 + 5)2 - 52 =
Katen [24]

C.171

Step-by-step explanation:

4×7(2)-25

4×49-25

multiply 4×49=196-25

then subtract 196-25=171

6 0
4 years ago
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Early diagnosis and treatment of malaria are necessary components in the control of malaria. The gold-standard light microscopy
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Answer:

a) 0.9548

b) 0.0452

Step-by-step explanation:

Specificity of test:

The specificity of a test is the true negative rate, that is, the proportion of people without the disease that test negative.

Sensitivity of test:

The sensitivity of a test is the true positive rate, which is the proportion of people with the disease that test positive.

a. What proportion of the people who test positive for Malaria would we expect to actually have the disease?

This is the sensitivity of the test, which is of 95.48%, so the proportion if 0.9548.

b. What proportion of people who test positive for Malaria will not have the disease?

True positive rate is of 95.48%.

1 - 0.9548 = 0.0452

So the proportion of people who test positive for Malaria will not have the disease is 0.0452.

6 0
3 years ago
Step 1: 5 (2x + 5) = 100
Triss [41]
He did not distribute the 5 to 5 so step two had to look like 10x+25=100
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3 years ago
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What is the quotient of 7^2/2x+6 divided by 3x-5/x+3
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\bf \cfrac{7^2}{2x+6}\div \cfrac{3x-5}{x+3}\implies \cfrac{7^2}{2x+6}\cdot \cfrac{x+3}{3x-5}\implies \cfrac{49}{2~~\begin{matrix} (x+3) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}}\cdot \cfrac{\begin{matrix} x+3 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix} }{3x-5}\implies \cfrac{49}{2(3x-5)}

6 0
4 years ago
Solve for z: z³-8=0 write your answers in simplified, rationlized form.​
tresset_1 [31]

Answer:

z=2

Step-by-step explanation:

4 0
4 years ago
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