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aleksklad [387]
3 years ago
14

Please answer the question attached in the image if you can use ms paint to draw arrows that will be appreciated.

Mathematics
1 answer:
yawa3891 [41]3 years ago
7 0
2 1 1/2 3 3 3/4 your welcome
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The answer is 83 ....

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The employees of a company were surveyed on questions regarding their educational background (college degree or no college degre
taurus [48]

Answer: D) 0.733.

Step-by-step explanation:

Let C denotes the number of employees having college degree and S denote the number of employees are single.

We are given ,

Total = 600 , n(C)=400  , n(S)=100 , n(C∩S)=60

Then,

n(C\cup S)=n(C)+n(S)-n(C\cap S)\\\\=400+100-60=440

Now, the probability that an employee of the company is single or has a college degree is

P(C\cup S)=\dfrac{n(C\cup S)}{\text{Total}}\\\\=\dfrac{440}{600}=0.733333333\approx0.733

Hence, the probability that an employee of the company is single or has a college degree is 0.733

8 0
3 years ago
ABC has vertices A(-4, 4), B(6, 0), and C(-4, 0). Is ABC a right triangle?
a_sh-v [17]

Check the picture below.

7 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
2 years ago
Solve for<br> 5(2 – 8) = 20
gladu [14]

Answer:

Step-by-step explanation:

5x - 40 = 20

5x = 20 + 40

5x = 60

x = 60/5

x = 12

6 0
2 years ago
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