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1 year ago

The length of a rectangle is six times its width. if the perimeter of the rectangle is 140 in , find its area.

1 answer:

4 0

L = 6W ---- the lenght is 6 times its width
P = 2(L + W) = 140
A =?

P = 2(L+W)
P = 2L + 2W
we know that L = 6W so we plug it into the equation
P = 2(6W) + 2W= 140
12W + 2W = 140
14W = 140
W = 140/14
W = 10

Lets plug 10 into the equation L = 6W
L = 6(10)
L = 60

AREA = LW
= 60 * 10
= 600

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E) 340 people participated in a conference. If the ratio of ladies and gentleman is

mariarad [96]

Answer:

<em>There were 204 ladies and 136 gentlemen in the conference</em>

Step-by-step explanation:

Let's set the variables:

L = number of ladies in the conference

G = number of gentlemen in the conference

We know 340 people participated in the conference, thus:

L + G = 340 [1]

We are also given the ratio of ladies and gentleman as 3:2, thus:

L/G = 3/2

Cross-multiplying:

2L = 3G [2]

From [1]:

L = 340 - G

Substituting in [2]:

2(340 - G) = 3G

Operating:

680 - 2G = 3G

5G = 680

G = 680/5 = 136

G = 136

L = 340 - 136 = 204

L = 204

There were 204 ladies and 136 gentlemen in the conference

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1 year ago

William is drafting his fantasy basketball team. He needs to select one player for each position. The following table shows how

vazorg [7]

Answer: 15120

Step-by-step explanation:

10×12×7×2×9=15120

3 0

1 year ago

Read 2 more answers

1. Mr. Villa bought 91.25 inches of plastic labeling tape. He will use 1.25 inches long to label each.How many labels can he mak

Cloud [144]

1.25x=91.25 now solve for x and we get x=73 so Mr. Villa can do 73 labels

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2 years ago

Which expression is equivalent to x Superscript negative five-thirds? StartFraction 1 Over RootIndex 5 StartRoot x cubed EndRoot

Anastasy [175]

Option B : $\frac{1}{\sqrt[3]{x^{5} } }$ is the expression equivalent to $x^{-\frac{5}{3}$

Explanation:

The given expression is $x^{-\frac{5}{3}$

Rewriting the expression $x^{-\frac{5}{3}$ using the exponent rule, $a^{-b}=\frac{1}{a^{b}}$

Hence, we get,

$\frac{1}{x^{\frac{5}{3} } }$

Simplifying, we get,

$\frac{1}{\left(x^{5}\right)^{\frac{1}{3}}}$

Applying the rule, $a^{\frac{1}{n}}=\sqrt[n]{a}$

Thus, we have,

$\frac{1}{\sqrt[3]{x^{5} } }$

Now, we shall determine from the options that which expression is equivalent to $x^{-\frac{5}{3}$

Option A: $\frac{1}{\sqrt[5]{x^{3} } }$

The expression $\frac{1}{\sqrt[5]{x^{3} } }$ is not equivalent to simplified expression $\frac{1}{\sqrt[3]{x^{5} } }$

Thus, the expression $\frac{1}{\sqrt[5]{x^{3} } }$ is not equivalent to $x^{-\frac{5}{3}$

Hence, Option A is not the correct answer.

Option B: $\frac{1}{\sqrt[3]{x^{5} } }$

The expression $\frac{1}{\sqrt[3]{x^{5} } }$ is equivalent to the simplified expression $\frac{1}{\sqrt[3]{x^{5} } }$

Thus, the expression $\frac{1}{\sqrt[3]{x^{5} } }$ is equivalent to $x^{-\frac{5}{3}$

Hence, Option B is the correct answer.

Option C: $-\sqrt[3]{x^5}$

The expression $-\sqrt[3]{x^5}$ is not equivalent to the simplified expression $\frac{1}{\sqrt[3]{x^{5} } }$

Thus, the expression $-\sqrt[3]{x^5}$ is not equivalent to $x^{-\frac{5}{3}$

Hence, Option C is not the correct answer.

Option D: $-\sqrt[5]{x^3}$

The expression $-\sqrt[5]{x^3}$ is not equivalent to the simplified expression $\frac{1}{\sqrt[3]{x^{5} } }$

Thus, the expression $-\sqrt[5]{x^3}$ is not equivalent to $x^{-\frac{5}{3}$

Hence, Option D is not the correct answer.

4 0

1 year ago

Read 2 more answers

12. If a manufacturer conducted a survey among randomly selected target market households and wanted to be 95% confident that th

atroni [7]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$

And rounded up we have that n=1068

Step-by-step explanation:

We have the following info given:

$Confidence= 0.95$ the confidence level desired

$ME =0.03$ represent the margin of error desired

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

The confidence level is 95% or 0.95, the significance is $\alpha=0.05$ and the critical value for this case using the normal standard distribution would be $z_{\alpha/2}=1.96$

Since we don't have prior information we can use $\hat p= 0.5$ as an unbiased estimator

Also we know that $ME =\pm 0.03$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$

And rounded up we have that n=1068

6 0

1 year ago