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2 years ago

Help! Which step is incorrect!

Mathematics

2 answers:

monitta2 years ago

8 0

Step (1) is incorrect. To divide by a number means to multiply by the reciprocal. In step (1) they did not convert the polynomial into its reciprocal.

SOVA2 [1]2 years ago

8 0

Answer:

step 1

Step-by-step explanation:

You are dividing. Therefore it has to be 1/6x-6

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Sandra is 4 feet tall. Pablo is 10% taller than Sandra and Michaels is 8% taller than Pablo. What is Michael height in feet

taurus [48]

We first calculate Pablo's height .

4 = 100%

x = 1% (Where x is 1% of Sandra's height.)

cross multiplying the equation, we get:

100x = 4 → x = 4/100 → x = 0.04

If 1% = 0.04, then 10% will be 0.04 × 10 = 0.4

Pablo's height is 4 + 0.4 = 4.4 feet.

Now we repeat the above steps to get Michal's height from Pablo's height

4.4 = 100%

x = 1% so:

100x = 4.4 → x = 4.4/ 100 → x = 0.044

If 1% = 0.044, then 8% will be 0.044 × 8 = 0.35

Michael's height is 4.4 + 0.35 = 4.75 feet.

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2 years ago

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hoa [83]

Answer:

b or c

Step-by-step explanation:

7 0

2 years ago

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30 treadmils to 36 elliptical machines

rewona [7]

Answer:

5/6 is the simplest form

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2 years ago

A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and

pav-90 [236]

Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean $\mu$ and standard deviation $\sigma$ , we have these following probabilities

$Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827$

$Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545$

$Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973$

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So $\mu = 53, \sigma = 11$

So:

$Pr(53-11 \leq X \leq 53+11) = 0.6827$

$Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545$

$Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973$

-----------

$Pr(42 \leq X \leq 64) = 0.6827$

$Pr(31 \leq X \leq 75) = 0.9545$

$Pr(20 \leq X \leq 86) = 0.9973$

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have $Pr(31 \leq X \leq 75) = 0.9545 = 0.9545$ subtracted by $Pr(42 \leq X \leq 64) = 0.6827$ is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.