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dolphi86 [110]
2 years ago
6

Which is the decimal expansion of the following 1/5

Mathematics
1 answer:
amid [387]2 years ago
7 0

Answer:

Decimal expansion of 1/5 = 0.2

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4 hundreds 6 hundredths - 3 hundredths <br> help
Tresset [83]

Answer:

7 hundredths

Step-by-step explanation:

add 4 to 6 and take away 3

4 0
3 years ago
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(2, 4) and (6, 12) find slope
Free_Kalibri [48]

Answer:

2

Step-by-step explanation:

Use the formula: change of y over change of x.

When you do that, your systems should look like this:

x1 = 2

y1 = 4

x2 = 6

y2 = 12

After that you plug these number into the corresponding variable.

(12 - 4) / (6 - 2) = 8/4 which equals 2.

4 0
3 years ago
Which term describes the variable y in the expression yn?<br> A.Product B.Factor C.Base D.Exponent
grigory [225]
If the expression is: yⁿ
then y is the Base.

If the expression is: yn
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5 0
3 years ago
Find the annual rate of interest. Principal = 4600 rupees, Period = 5 years, Total amount = 6440 rupees, Annual rate of interest
Len [333]

The annual rate of interest per year is 8%

<u>Solution:</u>

Given:- Principal (p) = 4600 rupees, Time –Period (t) = 5 years, Total amount(A) = 6440 rupees

First we will calculate the Interest and then using formula of simple interest we will calculate the rate of interest

Interest = Amount – Principal

Interest = 6440  – 4600 = 1840

Now using the formula of simple Interest and on putting values we get,

\text {Simple Interest }=\frac{P \times R \times T}{100}

Where "P" is the principal and "R" is the rate of interest per annum and "T" is the time period

1840=\frac{4600 \times R \times 5}{100}

\begin{aligned} \mathrm{R} &=\frac{1840 \times 100}{4600 \times 5} \\\\ \mathrm{R} &=\frac{1840}{230} \\\\ \mathrm{R} &=8 \% \end{aligned}

Hence, the required rate of interest per year is 8%

6 0
3 years ago
Suppose you are determining whether the depth of groves cut into aluminum by a milling machine is equal to 1.7 mm. You cut 14 gr
Gemiola [76]

Answer:

For this case we want to check if the true mean for the depth of groves cut into aluminium by a machine is equal to 1.7 (null hypothesis) and the alternative hypothesis would be the complement different from 1.7. And the best system of hypothesis are:

Null hypothesis: \mu =1.7

Alternative hypothesis \mu \neq 1.7[/tx]And the best system of hypothesis are:3. This two-sided test:&#10;H0: μ = 1.7 mm&#10;H1: μ ≠ 1.7 mmStep-by-step explanation:For this case we want to check if the true mean for the depth of groves cut into aluminium by a machine is equal to 1.7 (null hypothesis) and the alternative hypothesis would be the complement different from 1.7. And the best system of hypothesis are:Null hypothesis: [tex]\mu =1.7

Alternative hypothesis [tex]\mu \neq 1.7[/tx]

And the best system of hypothesis are:

3. This two-sided test:

H0: μ = 1.7 mm

H1: μ ≠ 1.7 mm

4 0
3 years ago
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