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snow_lady [41]
3 years ago
8

Divide 85/10 and 11/6​

Mathematics
2 answers:
Gnesinka [82]3 years ago
7 0

Answer:

85/10 = 8.5

11/6 = 1.83333333

Step-by-step explanation:

Step2247 [10]3 years ago
5 0

Step-by-step explanation:

change the ÷ into × flip the number too.

\frac{85}{10}  \div  \frac{11}{6}  =  \frac{85}{10}  \times  \frac{6}{11}

Reduce the fraction with 5

\frac{85}{10}  \times  \frac{6}{11}  =  \frac{17}{2}  \times  \frac{6}{11}

Reduce the numbers with the greatest common factor 2

\frac{17}{2}  \times  \frac{6}{11}  =  \frac{17}{1}  \times  \frac{3}{11}

Just multiply now, To multiply the fractions , multiply the numerators and denominators separately

\frac{17 \times 3}{1 \times 11}  = \frac{51}{11}

Solution :

\frac{51}{11}

Alternative form:

4 \frac{7}{11}

4.63

your welcome :) Mark me brainliest

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Kent multiplies both sides of the equation below by an expression. k startfraction 12 over k endfraction = 8 then he moves all t
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The resulting value of k is 12/7.

<h3>What is a fraction?</h3>

Fractions are the numerical values that are a part of the whole. A whole can be an object or a group of objects.

The steps below are presented in order to arrive at the value of k of the given equation.

First, multiply both sides of the equation by the variable k since the left-hand side of the equation has it in the denominator.

The solution of the equation will be;

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Hence, the resulting value of k is 12/7.

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2 years ago
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The life span of a domestic cat is normally distributed with a mean of 15.7 years and a standard deviation of 1.6 years. a. Find
Anvisha [2.4K]

Answer

a. 0.856

b. 0.78071

c. It is not unusual

d. 13.65 years old

Step-by-step explanation:

The life span of a domestic cat is normally distributed with a mean of 15.7 years and a standard deviation of 1.6 years.

We solve this question using z score formula:

z = (x-μ)/σ, where

x is the raw score

μ is the population mean

σ is the population standard deviation.

a. Find the probability that a cat will live to be older than 14 years.

For x > 14 years

z = 14 - 15.7/1.6

z = -1.0625

Probability value from Z-Table:

P(x<14) = 0.144

P(x>14) = 1 - P(x<14) = 0.856

b. Find the probability that a cat will live between 14 and 18 years.

For x = 14 years

z = 14 - 15.7/1.6

z = -1.0625

Probability value from Z-Table:

P(x = 14) = 0.144

For x = 18 years

z = 18 - 15.7/1.6

z= 1.4375

Probability value from Z-Table:

P(x = 18) = 0.92471

The probability that a cat will live between 14 and 18 years is calculated as:

P(x = 18) - P(x = 14)

0.92471 - 0.144

= 0.78071

c. If a cat lives to be over 18 years, would that be unusual? Why or why not?

For x > 18 years

z = 18 - 15.7/1.6

z= 1.4375

Probability value from Z-Table:

P(x<18) = 0.92471

P(x>18) = 1 - P(x<18) = 0.075288

Converting this to percentage:

0.075288 × 100 = 7.5288%

Hence, 7.5288% of the cats live to be over 18 years. Hence, it is not unusual.

d. How old would a cat have to be to be older than 90% of other cats?

From the question above, 10% of the cats would be older than 90% of other cats.

Hence, we find the z score of the 10th percentile

= -1.282

Hence,

-1.282 = x - 15.7/1.6

Cross Multiply

-1.282 × 1.6 = x - 15.7

- 2.0512 = x - 15.7

x = 15.7 -2.0512

x = 13.6488 years old

Approximately = 13.65 years old

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Answer:

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Monthly payment = $ 657.80

Total amount to be paid in a year = $ 657.80 × 12 = $ 7893.6

Interest amount in 1 year = $ 7893.6 - $ 7250 =  $ 643.6.

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