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11Alexandr11 [23.1K]
3 years ago
9

TIMED TEST WILL MARK BRAINLIEST PLEASEEE HELP

Mathematics
1 answer:
Virty [35]3 years ago
4 0
The answer is 15 hope this helps!
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Standard form 8 hundreds 8 thousands​
morpeh [17]

Answer:

omggggg

Step-by-step explanation:

3 0
3 years ago
What is -2b - 6 + 3b - 2 simplified?
Naya [18.7K]

Answer:

b - 8

Step-by-step explanation:

3b - 2b = b

-6 - 2 = 8

= b - 8

5 0
4 years ago
Read 2 more answers
Wht is the value of x in the proportion 2 1/4 ÷ x = 1 1/4 ÷ 3 3/5
balandron [24]

Answer:

x = \frac{162}{25}

Step-by-step explanation:

You need to solve the equation for x.

First change all the mixed numbers to improper fractions. Remember that a improper fraction the numerator (top number) is larger than the denominator (bottom number).  

2\frac{1}{4} / x = 1\frac{1}{4} / 3\frac{3}{5}\\\

All you need to do is multiply the numerator with the whole number, then add to the numerator:

\frac{9}{4} / x = \frac{5}{4} / \frac{18}{5}

Now simplify, to get rid of the numbers that are dividing:

(\frac{9}{4} ) (x)/(( x) (\frac{5}{18})) = (\frac{5}{4})(x) / ((\frac{18}{5})( \frac{5}{18} )

Cancel same numbers that are multiplying and dividing:

\frac{9}{4} / \frac{5}{18}  = \frac{5}{4} (x)

Clear X:

x = (\frac{9}{4}) (\frac{4}{5} )/ \frac{5}{18}

Simplify

x = ( \frac{9}{5} ) ( \frac{18}{5} )

you will get the following improper fraction: \frac{162}{25\\}

and for the mixed fraction you will have 6\frac{12}{25}

8 0
4 years ago
g In R simulate a sample of size 20 from a normal distribution with mean µ = 50 and standard deviation σ = 6. Hint: Use rnorm(20
Illusion [34]

Answer:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

Step-by-step explanation:

For this case first we need to create the sample of size 20 for the following distribution:

X\sim N(\mu = 50, \sigma =6)

And we can use the following code: rnorm(20,50,6) and we got this output:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

5 0
3 years ago
Find the length of f
JulijaS [17]
The proportion is 1.75
3 0
3 years ago
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