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Goryan [66]
4 years ago
12

Marisol is preparing a care package to send her brother the package will include a board game that weighs 4 lb in several 1/4 lb

snack packs the total weight of the care package must be less than 25 pounds how many snack packs can she send
Mathematics
2 answers:
alexgriva [62]4 years ago
8 0
25-4= 21

1/4=.25

21/.25= 84

You want it to be less than 25 pounds, so I would say:

Marisol can send 83 snack packs. 

I hope this helps!
~kaikers


igomit [66]4 years ago
3 0
1/4 =.25
25 - 4 = 21
x = packs of snacks
21 = x(.25)
21 = 84×.20
so he can send 84 pack of snacks.
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Peter, henry and john shared $345 between them. Peter received $45, henry received $75 and john received the rest.
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Answer:

Step-by-step explanation:

Let J = John's share ($)

     P = Peter's share ($)

     H = Henry's share ($)

 

1) H = P + 80

 

2) J/P = 6/3 = 2

 

3) P/H = 3/5  

 

Substitute equation 1) into equation 3)

 

P/(P + 80) = 3/5            (multiply both sides by (P + 80)

 

P = (3/5)*(P + 80)         (multiply both sides by 5)

 

5*P = 3*P + 240

 

2*P = 240

 

P = 240/2 = $120

 

Substitute P value into equation 2)

 

J/P = 2  or

 

J = 2*P = 2*$120 = $240

 

John received $240

 

 

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Suppose that prices of a certain model of new homes are normally distributed with a mean of $150,000. use the 68-95-99.7 rule to
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Answer:

We want to find the percentage of values between 147700 and 152300

P(147700

And one way to solve this is use a formula called z score in order to find the number of deviations from the mean for the limits given:

z= \frac{x-\mu}{\sigma}

And replacing we got:

z=\frac{147700-150000}{2300}=-1

z=\frac{152300-150000}{2300}=1

So then we are within 1 deviation from the mean so then we can conclude that the percentage of values between $147,700 and $152,300 is 68%

Step-by-step explanation:

We define the random variable representing the prices of a certain model as X and the distirbution for this random variable is given by:

X \sim N(\mu = 150000, \sigma =2300

The empirical rule states that within one deviation from the mean we have 68% of the data, within 2 deviations from the mean we have 95% and within 3 deviations 99.7 % of the data.

We want to find the percentage of values between 147700 and 152300

P(147700

And one way to solve this is use a formula called z score in order to find the number of deviations from the mean for the limits given:

z= \frac{x-\mu}{\sigma}

And replacing we got:

z=\frac{147700-150000}{2300}=-1

z=\frac{152300-150000}{2300}=1

So then we are within 1 deviation from the mean so then we can conclude that the percentage of values between $147,700 and $152,300 is 68%

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