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2 years ago

Circle A and circle B are congruent. CD is a chord of both circles. If AB = 8 ft and CD = 6 ft, how long is a radius?

1 answer:

7 0

Let us say that the intersection point of lines AB and CD is called point E. The lines AB and CD are perpendicular to each other which also means that the triangle CEB is a right triangle.

Where the line CB is the radius of the circle while the side lengths are half of the whole line segment:

EB = 0.5 AB = 0.5 (8 ft) = 4 ft

CE = 0.5 CD = 0.5 (6 ft) = 3 ft

Now using the hypotenuse formula since the triangle is right triangle, we can find for the radius or line CB:

CB^2 = EB^2 + CE^2

CB^2 = (4 ft)^2 + (3 ft)^2

CB^2 = 16 ft^2 + 9 ft^2

CB^2 = 25 ft^2

<span>CB = 5 ft = radius</span>

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The physician tells a woman who usually drinks 5 cups of coffee daily to cut down on her coffee consumption by 75%. If this woma

Maru [420]

Answer:

Step-by-step explanation:

5 cups per day.....1 cup = 8oz.....so 5 cups = (5 * 8) = 40 oz

cut down by 75%....means ur only drinking 25%

25% of 40 oz =

0.25(40) = 10 oz per day <===

6 0

2 years ago

A sample of 900 college freshmen were randomly selected for a national survey. Among the survey participants, 372 students were

Rufina [12.5K]

Answer:

a) $ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423$

b) $0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371$

Step-by-step explanation:

1) Data given and notation

n=900 represent the random sample taken

X=372 represent the students were pursuing liberal arts degrees

$\hat p=\frac{372}{900}=0.413$ estimated proportion of students were pursuing liberal arts degrees

$\alpha=0.01$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

p= population proportion of students were pursuing liberal arts degrees

2) Solution to the problem

The confidence interval would be given by this formula

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

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

The margin of error is given by:

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

If we replace we have:

$ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423$

And replacing into the confidence interval formula we got:

$0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371$

$0.413 + 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.455$

And the 99% confidence interval would be given (0.371;0.455).

We are confident (99%) that about 37.1% to 45.5% of students were pursuing liberal arts degrees.

7 0

1 year ago

The earth has a mass of approximately 6\cdot 10^{24}6⋅10 24 6, dot, 10, start superscript, 24, end superscript kilograms (\text{

Alex_Xolod [135]

Answer:

0.02

Step-by-step explanation:

The volume of the earth's oceans is approximately 1.34\cdot 10^{9}1.34⋅10

1, point, 34, dot, 10, start superscript, 9, end superscript cubic kilometers (\text{km}^3)(km

)left parenthesis, start text, k, m, end text, cubed, right parenthesis, and ocean water has a mass of about 1.03 \cdot 10^{12}\,\dfrac{\text{kg}}{\text{km}^3}1.03⋅10

1, point, 03, dot, 10, start superscript, 12, end superscript, start fraction, start text, k, g, end text, divided by, start text, k, m, end text, cubed, end fraction .

To simplify, we will use the product of powers property of exponents that says that x^a\cdot x^b = x^{a+b}x

⋅x

a+b

x, start superscript, a, end superscript, dot, x, start superscript, b, end superscript, equals, x, start superscript, a, plus, b, end superscript.

\qquad 1.34\cdot 10^{9}\,\cancel{\text{km}^3} \cdot 1.03 \cdot 10^{12}\,\dfrac{\text{kg}}{\cancel{\text{km}^3}} = 1.3802 \cdot 10^{21}\,\text{kg}1.34⋅10

⋅1.03⋅10

=1.3802⋅10

kg1, point, 34, dot, 10, start superscript, 9, end superscript, start cancel, start text, k, m, end text, cubed, end cancel, dot, 1, point, 03, dot, 10, start superscript, 12, end superscript, start fraction, start text, k, g, end text, divided by, start cancel, start text, k, m, end text, cubed, end cancel, end fraction, equals, 1, point, 3802, dot, 10, start superscript, 21, end superscript, start text, k, g, end text

Hint #2

Next we want to know what portion of the earth's mass this represents. We have:

\qquad \begin{aligned} \dfrac{\text{mass of the oceans}}{\text{total mass of the earth}} &= \dfrac{1.3802 \cdot 10^{21}\,\text{kg}}{6\cdot 10^{24}\,\text{kg}} \\\\ &= \dfrac{1.3802}{6\cdot 10^{3}} \\\\ &= \dfrac{1.3802}{6000} \\\\ &= 0.0002300\overline{3} \end{aligned}

total mass of the earth

mass of the oceans

6⋅10

1.3802⋅10

6⋅10

1.3802

6000

1.3802

=0.0002300

To convert this to a percent, we multiply by 100100100, so the oceans represent 0.02300\overline{3}\%0.02300

%0, point, 02300, start overline, 3, end overline, percent of the earth's total mass, according to these figures.

Hint #3

To the nearest hundredth of a percent, 0.020.020, point, 02 percent of the earth's mass is from oceans.

6 0

1 year ago

Consider the following formula used to estimate height, h, of a seedling after w weeks since planting:

svp [43]

Answer:

Option B. $w=5h-8$

Step-by-step explanation:

we have

$h=(\frac{1}{5})w+(\frac{8}{5})$

Solve for w

That means ----> isolate the variable w

Multiply by 5 both sides to remove the fraction

$5h=w+8$

subtract 8 both sides

$5h-8=w+8-8$

$5h-8=w$

Rewrite

$w=5h-8$

4 0

1 year ago

Kendra has a bowl containing 10 blue marbles, 15 yellow marbles, 5 green marbles, and 6 red marbles of the same size. She will r

Mice21 [21]

Given:
10 blue marbles ; 15 yellow marbles, 5 green marbles, 6 red marbles
A total of 36 marbles.

Probabilities:
1st marble: red = 6/36
2nd marble: yellow = 15/35

6/36 * 15/35 = 6*15 / 36*35 = 90/1260 = 1/14

The probability that Kendra will draw a red marble and then a yellow marble is 1/14.

6 0

2 years ago