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777dan777 [17]
3 years ago
13

Gina wants to ship 3 books that weigh 2 7/16 pounds, 1 7/8 pounds and 1/2 pound. the maximum weight she can ship is 6 pounds. ho

w much did Gina ship?
Mathematics
2 answers:
pshichka [43]3 years ago
6 0
You have to find a common denominator 1st then add. So 16 is the common denominator the problem should be 2 7/16+1 14/16+8-16= 3 29/16 then you need to divide 29/16 you will have 4 13/16
FrozenT [24]3 years ago
3 0

Answer:

Gina can ship all three books.        

Step-by-step explanation:

Given : Gina wants to ship 3 books that weigh 2 7/16 pounds, 1 7/8 pounds and 1/2 pound. The maximum weight she can ship is 6 pounds.

To find : How much did Gina ship?

Solution :

First we convert the mixed fraction into simpler fraction,

First book weight 2\frac{7}{16}=\frac{39}{16} pounds.

Second book weight 1\frac{7}{8}=\frac{15}{8} pounds.

Third book weight \frac{1}{2} pounds.

Total weight of books Gina have,

T=\frac{39}{16}+\frac{15}{8}+\frac{1}{2}

T=\frac{39+30+8}{16}

T=\frac{77}{16}

T=4\frac{13}{16}

T=4.81

4.81<6 pounds.

Which means she can ship all 3 books.

Therefore, Gina can ship all three books.

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no

Step-by-step explanation:

false

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8 0
3 years ago
Adds up to 13 multiplies to -48
fiasKO [112]
You only need two numbers to do that.
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7 0
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wolverine [178]

Answer:

y^6

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lbvjy [14]

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Step-by-step explanation:

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Read 2 more answers
According to a recent​ survey, the population distribution of number of years of education for​ self-employed individuals in a c
creativ13 [48]

Answer:

a) X: number of years of education

b) Sample mean = 13.5, Sample standard deviation = 0.4

c) Sample mean = 13.5, Sample standard deviation = 0.2

d) Decrease the sample standard deviation

Step-by-step explanation:

We are given the following in the question:

Mean, μ = 13.5 years

Standard deviation,σ = 2.8 years

a) random variable X

X: number of years of education

Central limit theorem:

If large random samples are drawn from population with mean \mu and standard deviation \sigma, then the distribution of sample mean will be normally distributed with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

b) mean and the standard for a random sample of size 49

\mu_{\bar{x}} = \mu = 13.5\\\\\sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{2.8}{\sqrt{49}} = 0.4

c) mean and the standard for a random sample of size 196

\mu_{\bar{x}} = \mu = 13.5\\\\\sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{2.8}{\sqrt{196}} = 0.2

d) Effect of increasing n

As the sample size increases, the standard error that is the sample standard deviation decreases. Thus, quadrupling sample size will half the standard deviation.

7 0
3 years ago
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