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Ymorist [56]
3 years ago
15

Help me please i’ll rate brainliest

Mathematics
1 answer:
Cloud [144]3 years ago
5 0

Answer:

<em>- 1.568</em>  

Step-by-step explanation:

Let's assume that the base of the given logarithm is 10.

log m = 0.345 ⇒ 10^{0.345} = m

log n = 1.223 ⇒ 10^{1.223} = n

mn = 10^{0.345} × 10^{1.223} = 10^{1.568}

\frac{1}{mn} = 10^{-1.568}

log 10^{-1.568} = - 1.568 × log 10 = <em>- 1.568</em>

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Elsa invested $500 in a fund for 4 years and was paid simple interest. The total interest that she received on the investment wa
dezoksy [38]

The annual interest rate of her investment would be = 4%

<h3>Calculation of annual interest rate</h3>

The principal amount invested= $500

The period of investment= 4years

The interest received= $80

The interest rate= ?

To calculate the interest rate use the formula,

SI = P×T×R/100

80 = 500 × 4 ×R/100

Make R the subject of formula,

R= 80 ×100/500×4

R= 8000/2000

R= 4%

Therefore, the annual interest rate of her investment would be = 4%.

Learn more about simple interest here:

brainly.com/question/20690803

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8 0
2 years ago
URGENT
Katyanochek1 [597]

Answer:

The answer to your question is Area = 62.5 u²

Step-by-step explanation:

Area = c \frac{a + b}{2}

a and b are bases

c = heigth

dAD = \sqrt{(x2 - x1)^{2} - (y2 - y1)^{2}  }

dAD = \sqrt{(-1 + 13)^{2} - (5 + 11)^{2}  }

dAD = \sqrt{(12)^{2} - (16)^{2}  }

dAD = \sqrt{144 + 256}

dAD = \sqrt{400}

dAD = 20 u

dBC = \sqrt{(3 - 0)^{2} - (2 + 2)^{2}  }

dBC = \sqrt{9 + 16}

dBC = \sqrt{25}

dBC = 5 u

dAB = \sqrt{(3 + 1)^{2} - (2 - 5)^{2}  }

dAB = \sqrt{(4)^{2} - (-3)^{2}  }

dAB = \sqrt{16 + 9}

dAB = \sqrt{25}

dAB = 5u

Area = 5 \frac{20 + 5}{2}

Area = 5 \frac{25}{2}

Area = 5(12.5)

Area = 62.5 u²

6 0
4 years ago
What is the value of b in this equation PLEASE HELP
Lina20 [59]

Answer:

d. 2

Step-by-step explanation:

see attached

7 0
3 years ago
4 x 7 =?
julia-pushkina [17]
The answer is a, i dont think i want the prize tho lol
6 0
3 years ago
Read 2 more answers
A machine used to fill​ gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of ounces and a st
11111nata11111 [884]

Complete question is;

A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 128 ounces and a standard deviation of 0.20 ounce. You randomly select 35 cans and carefully measure the contents. The sample mean of the cans is 127.9 ounces. Does the machine need to be? reset? Explain your reasoning.

(yes/no)?, it is (very unlikely/ likely) that you would have randomly sampled 35 cans with a mean equal to 127.9 ?ounces, because it (lies/ does not lie) within the range of a usual? event, namely within (1 standard deviation, 2 standard deviations 3 standard deviations) of the mean of the sample means.

Answer:

Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.

Step-by-step explanation:

We are given;

Mean: μ = 128

Standard deviation; σ = 0.2

n = 35

Now, formula for standard error of mean is given as;

se = σ/√n

se = 0.2/√35

se = 0.0338

Normally, the range of values should be within 2 standard deviations of mean. In this case, normal range of values will be;

μ ± 2se = 128 ± 0.0338

This gives; 127.9662, 128.0338

So, Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.

4 0
3 years ago
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