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Lena [83]
3 years ago
9

Someone please help me

Mathematics
1 answer:
Maurinko [17]3 years ago
8 0
The correct answer would be 13:4. <em>
</em>
<em />When reducing, you can divide both 600 and 1950 by 150. 1950/ 150= 13 and 600/150=4.<em>
</em>
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erica [24]

Answer:

f=u+2

Step-by-step explanation:

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3 years ago
Write three equivalent ratios to the given<br> ratios 4 to 8 and 5 to 10?
oee [108]

Answer:

1 to 2, 2 to 4, 3 to 6

Step-by-step explanation:

4s double is 8, 5s double is 10, so whichever you choose just double it.

In mathematically, 8/4 = 2, 10/5 = 2, so x to 2x

8 0
2 years ago
1. If the temperature drops from 13.5°F to -13.5°F, what is the change in temperature?
mrs_skeptik [129]
The change of temperature is 27 and the perimeter is 324
6 0
3 years ago
I need health drink is 130% of the recommended daily allowance RDA for a certain vitamin the RDA for this vitamin is 45 MG. How
Lerok [7]

Answer:

58.5mg of the vitamin in a drink

Step-by-step explanation:

We know...

130% of RDA for vitamin x >> RDA for vitamin x = 45mg >> 130% of 45 = ?? mg in a drink.

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I hope this helps you?!

3 0
3 years ago
Read 2 more answers
The proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control a
Dafna11 [192]

Answer:

Probability that at least 490 do not result in birth defects = 0.1076

Step-by-step explanation:

Given - The proportion of U.S. births that result in a birth defect is approximately 1/33 according to the Centers for Disease Control and Prevention (CDC). A local hospital randomly selects five births and lets the random variable X count the number not resulting in a defect. Assume the births are independent.

To find - If 500 births were observed rather than only 5, what is the approximate probability that at least 490 do not result in birth defects

Proof -

Given that,

P(birth that result in a birth defect) = 1/33

P(birth that not result in a birth defect) = 1 - 1/33 = 32/33

Now,

Given that, n = 500

X = Number of birth that does not result in birth defects

Now,

P(X ≥ 490) = \sum\limits^{500}_{x=490} {^{500} C_{x} } (\frac{32}{33} )^{x} (\frac{1}{33} )^{500-x}

                 = {^{500} C_{490} } (\frac{32}{33} )^{490} (\frac{1}{33} )^{500-490}  + .......+ {^{500} C_{500} } (\frac{32}{33} )^{500} (\frac{1}{33} )^{500-500}

                = 0.04541 + ......+0.0000002079

                = 0.1076

⇒Probability that at least 490 do not result in birth defects = 0.1076

4 0
3 years ago
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