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melomori [17]
3 years ago
13

What number is 4 times as many as 25

Mathematics
2 answers:
Sonbull [250]3 years ago
6 0
100. 25 4 times is 100. like 25 added 4 times 


prohojiy [21]3 years ago
3 0
4 times as many as 25 is 100

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Math question down below
e-lub [12.9K]
Since the absolute value of any number is always positive, the absolute value of -2 2/3 is 2 2/3.

Hope this helps!
3 0
3 years ago
solve each equation. Round your answers to the nearest ten-thousandth. Show your work please. Part 2A​
Reika [66]

Answer:

16. x = 3.0822, -3.0822

17. x = 1.6432, -1.6432

18. x = 1, -1

Step-by-step explanation:

16. log5[(4x²+2)/8] = 1

(4x²+2)/8 = 5¹

4x² + 2 = 40

4x² = 38

x² = 38/4

x² = 19/2

x = +/- sqrt(19/2)

x = 3.0822, -3.0822

17. log7[(2x²-4)(5)] = 1

10x² - 20 = 7¹

10x² = 27

x² = 2.7

x = +/- sqrt(2.7)

x = 1.6432, -1.6432

18. log5[(2x²+8)/2] = 1

x² + 4 = 5¹

x² = 1

x = +/-1

4 0
3 years ago
If a student places in the 99th percentile on an exam, she performed better than 99% of all students who completed the exam. Her
anastassius [24]

Answer:

Cumulative frequency distribution

Step-by-step explanation:

Cumulative frequency distribution is a form of a frequency distribution that represents the sum of a class and all classes below it. Remember that frequency distribution is an overview of all distinct values (or classes of values) and their respective number of occurrences.

3 0
3 years ago
A car rental agency charges $16.00 per day plus $0.15 per mile. Jim's bill for 4 days was $79.00. How many miles did he drive?
Ostrovityanka [42]
The equation would be: 16(4)+.15m=79
The answer would be 100 miles.
8 0
3 years ago
Read 2 more answers
discrete random variable X has the following probability distribution: x 13 18 20 24 27 P ( x ) 0.22 0.25 0.20 0.17 0.16 Compute
tatuchka [14]

Answer:

(a) P(X = 18) = 0.25

(b) P(X > 18) = 0.53

(c) P(X ≤ 18) = 0.47

(d) Mean = 19.76

(e) Variance = 22.2824

(f) Standard deviation = 4.7204

Step-by-step explanation:

We are given that discrete random variable X has the following probability distribution:

            X                    P (x)             X * P(x)            X^{2}             X^{2} * P(x)

           13                    0.22              2.86              169              37.18

           18                    0.25              4.5                324               81

           20                   0.20               4                  400               80

           24                    0.17              4.08              576              97.92

           27                    0.16              4.32              729             116.64

(a) P ( X = 18) = P(x) corresponding to X = 18 i.e. 0.25

     Therefore, P(X = 18) = 0.25

(b) P(X > 18) = 1 - P(X = 18) - P(X = 13) = 1 - 0.25 - 0.22 = 0.53

(c) P(X <= 18) = P(X = 13) + P(X = 18) = 0.22 + 0.25 = 0.47

(d) Mean of X, \mu = ∑X * P(x) ÷ ∑P(x) = (2.86 + 4.5 + 4 + 4.08 + 4.32) ÷ 1

                                                         = 19.76

(e) Variance of X, \sigma^{2} = ∑X^{2} * P(x) - (\sum X * P(x))^{2}

                                 = 412.74 - 19.76^{2} = 22.2824

(f) Standard deviation of X, \sigma = \sqrt{variance} = \sqrt{22.2824} = 4.7204 .

8 0
3 years ago
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