3 years ago

Match the equation with the graph -0.4x-0.8y=0.10

Mathematics

1 answer:

laiz [17]3 years ago

7 0

Can't see the graph, but it should look something like this:

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If you can help with as many as you can that would be greatly appreciated.

Solnce55 [7]

9. 0.26666
10. 2.375
11. 7/20
12. 206111/<span>50000</span>

6 0

4 years ago

Please help and I need EXPLANATION too

lys-0071 [83]

a because

$x + 1 \geqslant 0 \\ \: \: \: - 1 \: \: \: \: - 1 \\ x \geqslant - 1$

4 0

3 years ago

Which of the following is one of the disadvantages of geothermal energy? *

S_A_V [24]

Answer:

None of above

Step-by-step explanation:

Mark brainliest

6 0

2 years ago

Anticipated consumer demand in a restaurant for free-range steaks next month can be modeled by a normal random variable with mea

QveST [7]

A.
$\mathbb P(X>1000)=\mathbb P\left(\dfrac{X-1200}{100}>\dfrac{1000-1200}{100}\right)=\mathbb P(Z>-2)$

Since about 95% of a normal distribution falls within two standard deviations of the mean, that leaves 5% that lie without, with 2.5% lying to either side.

$\mathbb P(Z>-2)=\mathbb P(|Z|2)=0.95+0.025=0.975$

b.
$\mathbb P(1100$

About 68% of a normal distribution lies within one standard deviation of the mean, so this probability is about 0.68.

c. You're looking for $k$ such that

$\mathbb P(X>k)=0.10$

Since

$\mathbb P(X>k)=\mathbb P\left(\dfrac{X-1200}{100}>\dfrac{k-1200}{100}\right)=\mathbb P(Z>k^*)=0.10$

occurs for $k^*\approx1.2816$ , it follows that

$\dfrac{k-1200}{100}=1.2816\implies k\approx1328$

So there's a probability of 0.10 for having a demand exceeding about 1328 pounds.

3 0

4 years ago

The area of the figure is about blank

alina1380 [7]

Answer:

6/2=3

3*10=30

10*3.14=31.4

30+31.4=61.4 or rounded to 61

3 0

3 years ago

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