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shusha [124]
3 years ago
8

Describe a real-world situation that can be modeled by the function y=2x-3.

Mathematics
1 answer:
Nana76 [90]3 years ago
5 0

Answer:

yeah, sounds purty good.

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Jenna's class brought in 40 pieces of fruit for a celebration and 32 pieces of fruit were eaten. What percentage of the fruit wa
Pepsi [2]

Answer:

20%

Step-by-step explanation:

40 pieces total

32 pieces eaten

40 - 32 = 8

8 pieces not eaten

Percentage = (part / whole) x 100%

Percentage = (8 / 40) x 100% = 0.2 x 100%

Percentage = 20%

4 0
2 years ago
Read 2 more answers
How to subtract a negative fraction from a positive fraction?
ikadub [295]
Let's take an example to illustrate this case:

<span>positive fraction = 2/7
</span>
<span>negative fraction = - 3/5

Now we need to subtract  </span><span>- 3/5 from 2/7

Right?

2/7 - (-3/5) = 2/7 + 3/5

here we need to unify the denominators as follows:

the lowest common factor between 7 and 5 is 35
2/7 = 10/35
3/5 = 21/35

Now back to </span><span>2/7 + 3/5:

</span><span><span>2/7 + 3/5 = 10/35 + 21/35 = (10+21)/35 = 31/35

That's it




Hope that helps you</span>
</span>
8 0
3 years ago
Write an equation with the restrictions x=14,x=2 and x=0
11Alexandr11 [23.1K]
-14, -2 and 0 opposite intergers
5 0
3 years ago
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
Can someone please help me
Vsevolod [243]
The restrictions for the equation is that the denominator can not be zero. So the restrictions for x would be what values make the denominator zero.

7x^2 + 6x = 0
factor
x(7x + 6) = 0
multiply any number by 0 and you get 0. So either x = 0 or 7x+6 = 0
since there's a x in the numerator x/x = 1 so this will not be a restriction. Then the only restriction is:
7x+6 = 0
7x = -6
x = -6/7
5 0
3 years ago
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