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kiruha [24]
3 years ago
5

Divide. Express your answer in simplest form. 8 5/12 ÷ 1 3/4?

Mathematics
2 answers:
nordsb [41]3 years ago
7 0
The answer for your question is 5/39
maksim [4K]3 years ago
3 0
ANSWER: I got 101/16
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Jim has a bank account balance of −$15.46. As soon as he realizes this, he deposits $25.50 in the account.
k0ka [10]

Answer:

$10.04

Step-by-step explanation:

-15.46+25.50

equals

25.50-15.46

4 0
3 years ago
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Translate to a proportion and solve.<br> 148.5 is 59 1/5% of​ what?
lesantik [10]

The actual value is 250.84 .

Formula

Percentage calculation =(given value/total value)*100

let the actual value is "x"

percentage we known is 59 1/5% (59.2%) and 59.2% of x is 148.5.

for the percentage,

x*59.2% =148.5

x*59.2/100=148.5

x=(148.5/59.2)*100

x=(2.5084)*100

x=250.84

The actual value is 250.84 .

Learn more about the percentage here:

brainly.com/question/14091961

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5 0
2 years ago
What is a three letter word used to show division in a word problem?
olasank [31]
A three-letter word used to show division in a word problem is PER.
6 0
4 years ago
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On Sarah's 16 GB flash drive, she can fit 16 video clips of about 1 GB each, two movies of about 8 GB each, or 4000 songs of abo
Leno4ka [110]

Answer:

She can fit either the 16 video clips or the two movies

Step-by-step explanation:

7 0
3 years ago
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Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 144 millimeters,
valentinak56 [21]

Answer:

The probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample  means will be approximately normally distributed.

Then, the mean of the distribution of sample mean is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample mean  is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

<em>μ</em> = 144 mm

<em>σ</em> = 7 mm

<em>n</em> = 50.

Since <em>n</em> = 50 > 30, the Central limit theorem can be applied to approximate the sampling distribution of sample mean.

\bar X\sim N(\mu_{\bar x}=144, \sigma_{\bar x}^{2}=0.98)

Compute the probability that the sample mean would differ from the population mean by more than 2.6 mm as follows:

P(\bar X-\mu_{\bar x}>2.6)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}} >\frac{2.6}{\sqrt{0.98}})

                           =P(Z>2.63)\\=1-P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

8 0
3 years ago
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