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jekas [21]
2 years ago
13

Which of the following is most likely the next step in the series? ​

Mathematics
1 answer:
kolbaska11 [484]2 years ago
7 0

Answer:

a

Step-by-step explanation:

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Simplify the fraction 84/96.
Fantom [35]

Answer:

7/8

Step-by-step explanation:

U have to use the highest common factor and the lowest common denominator

3 0
3 years ago
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Write an equation in slope-intercept form for the line with slope 1/2 and y-intercept -9
IgorLugansk [536]

Answer:

y = 1/2x - 9

4 0
4 years ago
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What is the place value of 2 in 23?
meriva
It's value is 20 :) in the tenths place
you can also use a place value chart :)
4 0
3 years ago
In the figure, lines a, b, and c are parallel and m24 = 48
Debora [2.8K]

Answer:

m<1 = 48°

m<2 = 132°

m<3 = 132°

m<5 = 48°

Step-by-step explanation:

Given:

m<4 = 48°

Required:

m<1, m<2, m<3, and m<5

Solution:

m<1 = m<4 (alternate exterior angles are congruent)

m<1 = 48° (substitution)

m<2 = 180° - (m<1) (linear pair theorem)

m<2 = 180° - 48° (substitution)

m<2 = 132°

m<3 = m<2 (alternate interior angles are congruent)

m<3 = 132° (substitution)

m<5 = m<4 (corresponding angles are congruent)

m<5 = 48° (substitution)

8 0
3 years ago
An honest die is rolled. If the roll comes out even (2, 4, or 6), you will win $1; if the roll comes out odd (1,3, or 5), you wi
jenyasd209 [6]

Answer:

(a) 50%

(b) 47.5%

(c) 2.5%

Step-by-step explanation:

According to the honest coin principle, if the random variable <em>X</em> denotes the number of heads in <em>n</em> tosses of an honest coin (<em>n</em> ≥ 30), then <em>X</em> has an approximately normal distribution with mean, \mu=\frac{n}{2} and standard deviation, \sigma=\frac{\sqrt{n}}{2}.

Here the number of tosses is, <em>n</em> = 2500.

Since <em>n</em> is too large, i.e. <em>n</em> = 2500 > 30, the random variable <em>X</em> follows a normal distribution.

The mean and standard deviation are:

\mu=\frac{n}{2}=\frac{2500}{2}=1250\\\\\sigma=\frac{\sqrt{n}}{2}=\frac{\sqrt{2500}}{2}=25

(a)

To not lose any money the even rolls has to be 1250 or more.

Since, <em>μ</em> = 1250 it implies that the 50th percentile is also 1250.

Thus, the probability that by the end of the evening you will not have lost any money is 50%.

(b)

If the number of "even rolls" is 1250, it implies that the percentile of 1250 is 50th.

Then for number of "even rolls" as 1300,

1300 = 1250 + 2 × 25

        = μ + 2σ

Then P (μ + 2σ) for a normally distributed data is 0.975.

⇒ 1300 is at the 97.5th percentile.

Then the area between 1250 and 1300 is:

Area = 97.5% - 50%

        = 47.5%

Thus, the probability that the number of "even rolls" will fall between 1250 and 1300 is 47.5%.

(c)

To win $100 or more the number of even rolls has to at least 1300.

From part (b) we now 1300 is the 97.5th percentile.

Then the probability that you will win $100 or more is:

P (Win $100 or more) = 100% - 97.5%

                                   = 2.5%.

Thus, the probability that you will win $100 or more is 2.5%.

7 0
3 years ago
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