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3 years ago

(9/5)^3 I have no idea

Mathematics

1 answer:

shtirl [24]3 years ago

7 0

Answer:

5.832

Step-by-step explanation:

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Target has a television on sale 20% off the regular price of $500.00. How much is the price after the markdown?

kolezko [41]

Answer:

What is 20 percent of 500.? = 100.

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Evaluate the equation for the indicated value, f(n)=5n+8 for f(4),

Colt1911 [192]

You would need to put 4 in for (n)so, 5(4)+8 then solve. 5 times 4=20 plus 8 is 28. I hope I could help!

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3 years ago

Lin's father is paying for a $20 meal. He has 20%-off coupon for the meal, After the discount, a 5% sales tax is applied. How mu

Serjik [45]

Answer:

$16.80

Step-by-step explanation:

20 % of 20 is 4 and you would take the 4 off for a discount. 5% of 16 is 0.8 and you would add it since it's an additional tax.

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3 years ago

A gambling book recommends the following "winning strategy"for the

pentagon [3]

Answer:

Please see attachment

Step-by-step explanation:

Please see attachment

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3 years ago

Gears produced by a grinding process are categorized either as conforming (suitable for their intended purpose), downgraded (uns

Svetach [21]

Answer:

A. P=0.4013

B. P=0.9884

C. P=0.3222

Step-by-step explanation:

We can model this with a binomial random variable.

The sample size is n=10.

The probability is p=0.05 for scrap and p=0.05+0.15=0.20 for degraded or scrap.

The probability of having k scrap gears in the sample is:

$P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{10}{k} 0.05^{k} 0.95^{10-k}\\\\\\$

The probability that one or more is scrap can be calculated as 100% less the probability that no one is scrap:

$P(x\geq1)=1-P(x=0)\\\\\\P(x=0) = \dbinom{10}{0} p^{0}(1-p)^{10}=1*1*0.5987=0.5987\\\\\\P(x\geq1)=1-0.5987=0.4013$

The probability that 8 or more are not scrap is equal to the probability of having 2 or less that are scrap:

$P(x\leq2)=P(x=0)+P(x=1)+P(x=2)\\\\\\P(x=0) = \dbinom{10}{0} p^{0}(1-p)^{10}=1*1*0.5987=0.5987\\\\\\P(x=1) = \dbinom{10}{1} p^{1}(1-p)^{9}=10*0.05*0.6302=0.3151\\\\\\P(x=2) = \dbinom{10}{2} p^{2}(1-p)^{8}=45*0.0025*0.6634=0.0746\\\\\\P(x\leq2)=0.5987+0.3151+0.0746=0.9884$

The probability that more than two are degraded or scrap (p=0.2) is calculated as:

$P(x>2)=1-P(x\leq2)=1-(P(x=0)+P(x=1)+P(x=2))\\\\\\P(x=0) = \dbinom{10}{0} p^{0}(1-p)^{10}=1*1*0.1074=0.1074\\\\\\P(x=1) = \dbinom{10}{1} p^{1}(1-p)^{9}=10*0.2*0.1342=0.2684\\\\\\P(x=2) = \dbinom{10}{2} p^{2}(1-p)^{8}=45*0.04*0.1678=0.3020\\\\\\P(x>2)=1-(0.1074+0.2684+0.3020)=1-0.6778=0.3222$

4 0

3 years ago