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2 years ago

18x^2+69x+60 Factor the expression

1 answer:

6 0

3(3x+4)(2x+5)

Because you factor out the 3 and then factor 6x^2+23x+20
That becomes (6x^2+8x)+(15x+20)
Factor out a two from the left and five from the right.
That makes 2x(3x+4) + 5(3x+4)
That becomes (3x+4)(2x+5)
Then put the three back in front.

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Max and Sharifa are both saving to buy the same model of DVD player, which costs $99, including tax. Max already has $31.00 and

gogolik [260]

Answer:

Sharifa can pay for the DVD player first.

Step-by-step explanation:

Max:

99 - 31 = $68 remaining

68 / 8.5 = 8 weeks

Sharifa:

99 - 25.5 = $73.50 remaining

73.5 / 10.5 = 7 weeks

7 0

2 years ago

7.1.35-T Question Help You are the operations manager for an airline and you are considering a higher fare level for passengers

shusha [124]

Answer:

You must survey 784 air passengers.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Assume that nothing is known about the percentage of passengers who prefer aisle seats.

This means that $\pi = 0.5$ , which is when the largest sample size will be needed.

Within 3.5 percentage points of the true population percentage.

We have to find n for which M = 0.035. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.035 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.035\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.035}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.035})^2$

$n = 784$

You must survey 784 air passengers.

6 0

2 years ago

The probability that a male will be colorblind is .042. Find the probabilities that in a group of 53 men, the following are true

Dvinal [7]

Answer:

See below for answers and explanations

Step-by-step explanation:

Part A

$\displaystyle P(X=x)=\binom{n}{x}p^xq^{n-x}\\\\P(X=5)=\binom{53}{5}(0.042)^5(1-0.042)^{53-5}\\\\P(X=5)=\frac{53!}{(53-5)!*5!}(0.042)^5(0.958)^{48}\\\\P(X=5)\approx0.0478$

Part B

$P(X\leq5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\\\\P(X\leq5)=\binom{53}{1}(0.042)^1(1-0.042)^{53-1}+\binom{53}{2}(0.042)^2(1-0.042)^{53-2}+\binom{53}{3}(0.042)^3(1-0.042)^{53-3}+\binom{53}{4}(0.042)^4(1-0.042)^{53-4}+\binom{53}{1}(0.042)^5(1-0.042)^{53-5}\\\\P(X\leq5)\approx0.9767$