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alex41 [277]
2 years ago
12

Chris wonders whether the planes that fly over his house follow any sort of pattern. For two months, he

Mathematics
1 answer:
KatRina [158]2 years ago
5 0

Answer:

Kindly check explanation

Step-by-step explanation:

From the relative frequency below:

For NORTH - SOUTH:

Monday - Thursday = 115 ; has a relative frequency of 75.16%,

Hence, we can obtain the row total since it amounts to 100% thus;

75.16% of row total = 115

0.7516× row total = 115

Row total = 115 / 0.7516 = 153.00

Hence, Friday - sunday:

Row total - (Monday-Thursday)

153 - 115 = 38

FOR EAST - WEST:

Monday - Thursday = 21 ; has a relative frequency of 25.30%,

Hence, we can obtain the row total since it amounts to 100% thus;

25.30% of row total = 21

0. 253 × row total = 21

Row total = 21 / 0.253 = 83.00

Hence, Friday - sunday:

Row total - (Monday-Thursday)

83 - 21 = 62

x = (38 / 100) × 100%

x = 0.38 × 100%

x = 38.00 %

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Step-by-step explanation:

the answer is 13.5


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At a local fitness center, members pay a $14 membership fee and $5 for each aerobics class. Nonmembers pay $6 for each aerobics
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Answer:

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1 year ago
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A quality control engineer tests the quality of produced computers. suppose that 5% of computers have defects, and defects occur
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(a) 0.059582148 probability of exactly 3 defective out of 20

 (b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.

  (a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So

 0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)

 = 0.05^3 * 0.95^17 * 20! / (3!17!)

 = 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)

 = 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)

 = 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)

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  (b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.

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3 0
2 years ago
You deposit $300 in a savings account that pays 6% interest compounded semiannually. How much will you have at the middle of the
Makovka662 [10]

Answer:

  • The total amount accrued, principal plus interest,  from compound interest on an original principal of  $ 300.00 at a rate of 6% per year  compounded 2 times per year  over 0.5 years is $ 309.00.

  • The total amount accrued, principal plus interest,  from compound interest on an original principal of  $ 300.00 at a rate of 6% per year  compounded 2 times per year  over 1 year is $ 318.27.

Step-by-step explanation:

a)  How much will you have at the middle of the first year?

Using the formula

A\:=\:P\left(1+\frac{r}{n}\right)^{nt}

where

  • Principle = P
  • Annual rate = r
  • Compound = n
  • Time  = (t in years)
  • A = Total amount

Given:

Principle P = $300

Annual rate r = 6% = 0.06 per year

Compound n = Semi-Annually = 2

Time (t in years) = 0.5 years

To determine:

Total amount = A = ?

Using the formula

A\:=\:P\left(1+\frac{r}{n}\right)^{nt}

substituting the values

A=300\left(1+\frac{0.06}{2}\right)^{\left(2\right)\left(0.5\right)}

A=300\cdot \frac{2.06}{2}

A=\frac{618}{2}

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Therefore, the total amount accrued, principal plus interest,  from compound interest on an original principal of  $ 300.00 at a rate of 6% per year  compounded 2 times per year  over 0.5 years is $ 309.00.

Part b) How much at the end of one year?

Using the formula

A\:=\:P\left(1+\frac{r}{n}\right)^{nt}

where

  • Principle = P
  • Annual rate = r
  • Compound = n
  • Time  = (t in years)
  • A = Total amount

Given:

Principle P = $300

Annual rate r = 6% = 0.06 per year

Compound n = Semi-Annually = 2

Time (t in years) = 1 years

To determine:

Total amount = A = ?

so using the formula

A\:=\:P\left(1+\frac{r}{n}\right)^{nt}

so substituting the values

A\:=\:300\left(1+\frac{0.06}{2}\right)^{\left(2\right)\left(1\right)}

A=300\cdot \frac{2.06^2}{2^2}

A=318.27 $

Therefore, the total amount accrued, principal plus interest,  from compound interest on an original principal of  $ 300.00 at a rate of 6% per year  compounded 2 times per year  over 1 year is $ 318.27.

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