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3 years ago

I need an answer to this

Mathematics

1 answer:

kogti [31]3 years ago

4 0

First, find the coordinates of all the points.

Then since we're dealing with a scale factor of 1/3 and we're assuming that it is about the origin, all you need to do is divide the coordinates by 3.

ex : (9,3) --> (3,1)

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Sally bought a bottle of laundry detergent that was 25% less than the bottle she bought last week, which was $9.60. How much is

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9.60 (.25) = 2.40
9.60 - 2.40 = 7.20
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3 years ago

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4 years ago

Which choice is a factor of 70 25,12,10, or 5

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6 0

3 years ago

Each Saturday morning, Janie works 5 hours and earns $35.75. How much does Janie earn for each hour she works?

BabaBlast [244]

Answer/Step-by-step explanation:

If Janie works 5 Hours and earn $35.75..

And It asking for Janie earn for each hour she works..

Therefore, we have to divide to Find how much Janie earn for each hour.

$35.75/5=$7.15

Hence. Janie earn $7.15 each Hour she work.

[RevyBreeze]

4 0

2 years ago

Read 2 more answers

According to an NRF survey conducted by BIGresearch, the average family spends about $237 on electronics (computers, cell phones

Usimov [2.4K]

Answer:

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is 0.0537.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is 0.0023.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is 0.1101.

Step-by-step explanation:

We are given that according to an NRF survey conducted by BIG research, the average family spends about $237 on electronics in back-to-college spending per student.

Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54.

Let X = back-to-college family spending on electronics

SO, X ~ Normal( $\mu=237,\sigma^{2} =54^{2}$ )

The z score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean family spending = $237

$\sigma$ = standard deviation = $54

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is = P(X < $150)

P(X < $150) = P( $\frac{X-\mu}{\sigma}$ < $\frac{150-237}{54}$ ) = P(Z < -1.61) = 1 - P(Z $\leq$ 1.61)

= 1 - 0.9463 = 0.0537

The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9463.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is = P(X > $390)

P(X > $390) = P( $\frac{X-\mu}{\sigma}$ > $\frac{390-237}{54}$ ) = P(Z > 2.83) = 1 - P(Z $\leq$ 2.83)

= 1 - 0.9977 = 0.0023

The above probability is calculated by looking at the value of x = 2.83 in the z table which has an area of 0.9977.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is given by = P($120 < X < $175)

P($120 < X < $175) = P(X < $175) - P(X $\leq$ $120)

P(X < $175) = P( $\frac{X-\mu}{\sigma}$ < $\frac{175-237}{54}$ ) = P(Z < -1.15) = 1 - P(Z $\leq$ 1.15)

= 1 - 0.8749 = 0.1251

P(X < $120) = P( $\frac{X-\mu}{\sigma}$ < $\frac{120-237}{54}$ ) = P(Z < -2.17) = 1 - P(Z $\leq$ 2.17)

= 1 - 0.9850 = 0.015

The above probability is calculated by looking at the value of x = 1.15 and x = 2.17 in the z table which has an area of 0.8749 and 0.9850 respectively.

Therefore, P($120 < X < $175) = 0.1251 - 0.015 = 0.1101

5 0

4 years ago