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2 years ago

The value of 8 in 56,982 is ___ times as large as the value of 8 in 156,408

Mathematics

2 answers:

ohaa [14]2 years ago

4 0

The value of 8 in 56982 is 10 times as large as the value of 8 in 156,408

Rudik [331]2 years ago

3 0

The value of 8 in 56982 is in the tens place value. The value of 8 in 156,408 is in the ones value

Divide

80/8 = 10

The value of 8 in 56,982 is 10 times as large as the value in 8 in 156,408

hope this helps

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Find the standard deviation of the following data. Answers are rounded to the nearest tenth. 5, 5, 6, 12, 13, 26, 37, 49, 51, 56

kogti [31]

Answer:

24.2(to the nearest tenth)

Step-by-step explanation:

The question is ungrouped data type

standard deviation =√ [∑ (x-μ)² / n]

mean (μ)=∑x/n

= $\frac{5+5+6+12+13+26+37+49+51+56+56+84}{12}$

=33.3

x-μ for data 5, 5, 6, 12, 13, 26, 37, 49, 51, 56, 56, 84 will be

-28.3, -28.3, -27.3, -21.3, -7.3, 3.7, 15.7,17.7, 22.7,22.7,50.7

(x-μ)² will be 800.89, 800.89,745.29,453.69,53.29,13.69,246.49,313.29,515.29,515.29,2570.49

∑ (x-μ)² will be = 7028.59

standard deviation = √(7028.59 / 12)

=24.2

6 0

2 years 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

Kathy has $50,000 to invest today and would like to determine whether it is realistic for her to achieve her goal of buying a ho

olga_2 [115]

Answer:

a) About 12%

Step-by-step explanation:

We need to find the interest rate required to achieve her goal, so we will need to use the interest-compound formula:

$FV=PV(1+i)^{n}$

Where:

PV= Present Value

i= interest rate

FV= Future Value

n= number of periods

replacing the data provided:

$150.000=50.000(1+i)^{10}$

solving for i:

first, divide both sides by 50.000 to simplify the equation:

$3=(1+i)^{10}$

Take $10^{th}$ roots of both sides:

$1+i=$ ± $\sqrt[10]{3}$

solve for i:

$i=$ ± $\sqrt[10]{3} -1$

We get two answers, but we look for a coherent value. So we take the positive one:

$i=11.6123174$ ≈12

6 0

2 years ago

Mr. Shamir employs two part-time typists, Inna and Jim, for his typing needs. Inna charges $15 an hour and can type 6 pages an h

coldgirl [10]

Answer:

The minimum cost would be 480$ when Inna works for 8 hours and Jim works for 20 hours.

Step-by-step explanation:

We are given the following information in the question:

Charges for 1 hour for Inna = $15

Number of pages typed by Inna in 1 hour = 6

Charges for 1 hour for Jim = $18

Number of pages typed by Jim in 1 hour = 8

Let x be the number of hours Inna work and let y be the number of hours Jim work.

Total cost = $15x + 18y$

We have to minimize this cost.

Then, we can write the following inequalities:

$6x + 8y \geq 208\\x \geq 8\\y \geq 8\\$

The corner points as evaluated from graph are: (8,20) and (24,8)

C(8,20) = 480$

C(24,8) = 504$

Hence, the minimum cost would be 480$ when Inna works for 8 hours and Jim works for 20 hours.

The attached image shows the graph.

3 0

2 years ago

A company administers an "aptitude test for managers" to aid in selecting new management trainees. Prior experience suggests tha

RoseWind [281]

Answer:

0.94

Step-by-step explanation:

The question after this basically is:

<em>"If the applicant passes the "aptitude test for managers", what is the probability that the applicant will succeed in the management position?"</em>

<em />

So,

P(successful if hired) = 60% = 0.6 [let it be P(x)]

P(success at passing the test) = 85% = 0.85 [let it be P(y)]

P(successful and pass the test) = P(x) + P(y) -[P(x)*P(y)]

So,

P(successful and pass the test) = 0.6 + 0.85 - (0.6*0.85) = 0.94 (94%)

5 0

1 year ago