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Inga [223]
3 years ago
12

2nd out of 10 questions

Mathematics
2 answers:
Fudgin [204]3 years ago
6 0

Assuming we're after the exact answer, we can just worry about the fractional parts

0.2 \times 0.3 \times 0.6 = 0.06 \times .6 = 0.036

Three decimal places, second choice

Lapatulllka [165]3 years ago
3 0

The multiplication will have a result with a number of decimal places equal to the sum of the numbers of decimal places in the factors: 1 + 1 + 1 = 3.

... 4.2 × 12.3 × 14.6 = 754.236

However, since this is a cost, it will be rounded to 2 decimal places.

The cost will be $754.24. . . . . . Paul's final answer

_____

Of the choices offered, the best one is

... three

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I'm not good at math so I need help with this
EleoNora [17]
1/2(6.3x10.4)=32.8m^2
5 0
3 years ago
Use distributive property to solve the equation 5( 2x +4) =30<br> A. x-2<br> B. x=1<br> c. =10
jeyben [28]

Answer:

B

Step-by-step explanation:

5(2x+4)=30

10x+20=30

10x=10

x=1

4 0
3 years ago
Read 2 more answers
QUESTION
yanalaym [24]

Answer:

$97,958.42

Step-by-step explanation:

To solve this problem we can use the compound interest formula which is shown below:

A=P(1+r )^{t}

<em>P = initial balance </em>

<em>r = interest rate </em>

<em>t = time </em>

<em> </em>

First change 6.5% to its decimal form:

6.5% -> \frac{6.5}{100} -> 0.065

Next plug in the values:

A=49,000(1+.065)^{11}

A=97,958.42

They have to pay back $97,958.42

7 0
3 years ago
Installments
ioda

Answer:

$ 338.70

Step-by-step explanation:

https://www.calculatorsoup.com/calculators/financial/loan-calculator-simple.php?pv=15%2C000.00&ratepercent=4&t=48&time_t=month&given_data_last=find_pmt&action=solve

6 0
3 years ago
Given a sample mean difference of 23, a standard error of 4.2, and a critical t value of 2.262, what are the values for the conf
erastova [34]

Answer: (13.4996, 32.5004)

Step-by-step explanation:

The confidence interval for mean difference is given by :-

(\overline{x}_1-\overline{x}_2)\pm t^*\times SE ,

where (\overline{x}_1-\overline{x}_2) = sample mean difference .

t^*= critical t value (two-tailed)

SE= Standard error.

Given : (\overline{x}_1-\overline{x}_2)=23

t^*=2.262

SE= 4.2

Now, the confidence interval for mean difference will be :-

23\pm (2.262)\times (4.2)\\\\= 23\pm9.5004\\\\=(23-9.5004,\ 23+9.5004)\\\\=(13.4996,\ 32.5004)

Hence, the required confidence interval : (13.4996, 32.5004)

8 0
3 years ago
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