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Lesechka [4]
3 years ago
9

Determine which fraction is larger 3/7 or 1/3 find the common denominator

Mathematics
1 answer:
Wewaii [24]3 years ago
8 0

Answer:

  • LCM is 21

{:}\leadsto\sf {\dfrac {3}{7}}

{:}\leadsto\sf {\dfrac {3×3}{7×3}}

{:}\leadsto\sf {\dfrac {9}{2}}

  • .

{:}\leadsto\sf {\dfrac {1}{3}}

{:}\leadsto\sf {\dfrac {1×7}{3×7}}

{:}\leadsto\sf {\dfrac {7}{21}}

<h2>____________________</h2>

{:}\leadsto\sf {\dfrac {9}{21}}\iff {\dfrac {7}{21}}

{:}\leadsto\sf {\dfrac{9}{21}} \gt {\dfrac {7}{21}}

\therefore\sf {\dfrac {3}{7}} \gt {\dfrac {7}{21}}

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Write the ratio for cosB
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Answer:

11/61

Step-by-step explanation:

cos b = adj side/hypotenuse

=11/61

4 0
2 years ago
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You have decided to invest $1000 in a savings bond that pays 4% interest, compounded semi-annually. What will the bond be worth
Evgen [1.6K]

Answer:

$2191.12

Step-by-step explanation:

We are asked to find the value of a bond after 10 years, if you invest $1000 in a savings bond that pays 4% interest, compounded semi-annually.

FV=C_0\times (1+r)^n, where,

C_0=\text{Initial amount},

r = Rate of return in decimal form.

n = Number of periods.

Since interest is compounded semi-annually, so 'n' will be 2 times 10 that is 20.

4\%=\frac{4}{100}=0.04

FV=\$1,000\times (1+0.04)^{20}

FV=\$1,000\times (1.04)^{20}

FV=\$1,000\times 2.1911231430334194

FV=\$2191.1231430334194

FV\approx \$2191.12

Therefore, the bond would be $2191.12 worth in 10 years.

3 0
3 years ago
Twice a number decreased by 22 is 48. What is the number?
klemol [59]

Answer:

Given - Twice a no. decrease by 22 is 48

To find - The no.

Solution -

Let the no. be X

According to the question, equation made is -

2x - 22 = 48

2x = 48 + 22

2x = 70

x = 70/2

X= 35

Verification -

35 ×2 - 22 = 48

70 - 22 = 48

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6 0
2 years ago
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An experiment was conducted to record the jumping distances of paper frogs made from construction paper. Based on the sample, th
Natasha2012 [34]

Answer:

9.9676 - 2.326*0.5904 =8.594

9.9676 + 2.326*0.5904 =11.341

Step-by-step explanation:

Notation

\bar X represent the sample mean

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

For this case the 9% confidence interval is given by:

8.8104 \leq \mu \leq 11.1248

We can calculate the mean with the following:

\bar X = \frac{8.8104 +11.1248}{2}= 9.9676

And we can find the margin of error with:

ME= \frac{11.1248- 8.8104}{2}= 1.1572

The margin of error for this case is given by:

ME = t_{\alpha/2}\frac{s}{\sqrt{n}} = t_{\alpha/2} SE

And we can solve for the standard error:

SE = \frac{ME}{t_{\alpha/2}}

The critical value for 95% confidence using the normal standard distribution is approximately 1.96 and replacing we got:

SE = \frac{1.1572}{1.96}= 0.5904

Now for the 98% confidence interval the significance is \alpha=1-0.98= 0.02 and \alpha/2 = 0.01 the critical value would be 2.326 and then the confidence interval would be:

9.9676 - 2.326*0.5904 =8.594

9.9676 + 2.326*0.5904 =11.341

8 0
2 years ago
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