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Elenna [48]
3 years ago
6

Simplify. 62+15 21 27 46 51

Mathematics
2 answers:
USPshnik [31]3 years ago
8 0
62+15 = 77, since it’s not an option i don’t know
Step2247 [10]3 years ago
3 0

Answer:The answer is 77.

Step-by-step explanation:

But it's not in the options.

If the question is 62-15=47

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Suppose that the following group of values has been entered into the TVM
Mila [183]

Answer:

The answer on A P E X is bal(132)

5 0
3 years ago
Solve.
bixtya [17]

Answer:

-3 ± 2\sqrt{3} = x

Step-by-step explanation:

x² + 6x + 9 = 12 → x² + 6x - 3

Since this Quadratic Expression is unfactorable, apply the <em>Quadratic</em><em> </em><em>Formula</em>:

x = -b ± \frac{\sqrt{{b}^{2} - 4ac}}{2a}

-6 ± \frac{\sqrt{{6}^{2} - 4(1)(-3)}}{2(1)} =   -6 ± \frac{\sqrt{36 + 12}}{2} = -6 ± \frac{\sqrt{48}}{2} \\ \\ -3 ± 2\sqrt{3}

√48 = √[3 × 16] = 4√3

½[4√3] = 2√3

Then attach half of -6 to it as well [-3]:

-3 ± 2√3

I am joyous to assist you anytime.

7 0
3 years ago
After being rejected for employment, Kim Kelly learns that the Bellevue Credit Company has hired only five women among the last
Georgia [21]

Answer:

The probability that 5 or fewer women are hired, assuming no gender discrimination, is 0.0317; we can use this result to support her charge of gender discrimination.

Step-by-step explanation:

If we are assuming that the women and the men are equally qualified, then the probability for each employee that is hired the probability for it to be a women should be 1/2. Note that the fact that more men that women are hired in a sample might not be disctrimination: for example, if 2 men are hired out of 2 employees, that can happen with probability 1/4, so it is quite common. In order to support her charge for gender discrimination, we need at least a probability less that 0.05 that 5 (or less) women are hired out of 19 employees.

Since each configuration is equally probable, we will count the total amount of possible cases that 5 or less women are hired, and dividide it by the total amount of cases, 2¹⁹.

  • 0 women hired: one possible case: every employee is male
  • 1 women hired: 19 possible cases
  • 2 women hired: {19 \choose 2} = 171 possible cases
  • 3 women hired: {19 \choose 3}  = 969 possible cases
  • 4 women hired: {19 \choose 4} = 3876 possible cases
  • 5 women hired: {19 \choose 5} = 11628 possible cases

Thus, there are a total of 11628+3876+969+171 = 16644 possible cases out of 2¹⁹ ones. All of them with seemingly equal probability. As a consequence, the probability of 5 or less women to be hired out of 19 employees, assuming that the probability to hire 1 is 1/2, is

16644/2¹⁹ = 0.0317 < 0.05

The probability that 5 or fewer women are hired, assuming no gender discrimination, is 0.0317. Since the probability is so low, we can conclude that for the employer, a woman equally qualified as a man is less likely to be hired, therefore, we can support her charge of gender discrimination.

6 0
3 years ago
A widget company produces 25 widgets a day,5 of which are defective.Find the probability of selecting 5 widgets from the 25 prod
MariettaO [177]

Answer:

0.292

Step-by-step explanation:

Combinations can be used to solve the following problem.

We are choosing 5 widgets from 25 = ²⁵C₅

We want to select zero widgets from defective widgets = ⁵C₀

From the 20 non-defective widgets we want to select 5 = ²⁰C₅

So the probability is:

P = ( ⁵C₀ * ²⁰C₅) / ²⁰C₅

P = (1 * 15504) / 53130

= 15504/ 53130

=0.292 ..

7 0
3 years ago
A study showed that 25% of the students drive themselves to school. Based on the suggested probability, in a class of 18 student
Gwar [14]

Answer:

B) 0.283

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

25% of the students drive themselves to school.

This means that p = 0.25

Class of 18 students

This means that n = 18

What would be the probability that at least 6 students drive themselves to school?

This is

P(X \geq 6) = 1 - P(X < 6)

In which

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{15,0}.(0.25)^{0}.(0.75)^{18} = 0.006

P(X = 1) = C_{15,1}.(0.25)^{1}.(0.75)^{17} = 0.034

P(X = 2) = C_{15,2}.(0.25)^{2}.(0.75)^{16} = 0.096

P(X = 3) = C_{15,3}.(0.25)^{3}.(0.75)^{15} = 0.17

P(X = 4) = C_{15,4}.(0.25)^{4}.(0.75)^{14} = 0.213

P(X = 5) = C_{15,5}.(0.25)^{5}.(0.75)^{13} = 0.199

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.006 + 0.034 + 0.096 + 0.17 + 0.213 + 0.199 = 0.718

P(X \geq 6) = 1 - P(X < 6) = 1 - 0.718 = 0.282

Closest option is B, just a small rounding difference.

4 0
3 years ago
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