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olya-2409 [2.1K]
2 years ago
7

HELP REAL FASTT PLEASE

Mathematics
2 answers:
PIT_PIT [208]2 years ago
4 0

Answer:

The forth one.

Step-by-step explanation:

I studied the things, and the forth one looks most accurate.

finlep [7]2 years ago
4 0
It’s possible that it is the fourth one
You might be interested in
3 1/8t=2 1/2 what is t?​
navik [9.2K]

Answer:

T=4

Step-by-step explanation:

it is just half of what the other side is if one side is like 1 1/2 then you just break the other side in half and that will give you T.

5 0
3 years ago
Number 7 I need help!
pshichka [43]

Answer:

  (1) is the appropriate choice

Step-by-step explanation:

Draw yourself a picture and see what happens.

Dilation does not change slopes; it only changes linear dimensions by the scale factor. A line five times as long with the same slope will match the description of selection (1).

3 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
How many grams are there in 3.3 x 10^23 atoms of boron?
iren [92.7K]

5.924 g

Step-by-step explanation:

The average weight of a boron atom is 10.811 atomic mass units (AMU). Therefore 3.3 x 10^23 atoms of boron will weight;

3.3 x 10^23 * 10.811 = 3.56763 x 10 ^ 24

One (1) gram = 6.022 x 1023 AMU. Therefore to convert AMU to grams;

3.56763 x 10 ^ 24 / 6.022 x 10^23

= 5.924 g

Learn More:

For more on atomic mass check out;

brainly.com/question/13058661

brainly.com/question/11014846

#LearnWithBrainly

8 0
3 years ago
I need help doing this can anyone help?
Olin [163]

You posted a lot of questions. I'll do the first four problems to get you started.

============================

Problem 1

This sequence is arithmetic because we add the same amount (4) each time to get the next term. The first term is -2. The first five terms are: -2, 2, 6, 10, 14

============================

Problem 2

This sequence is geometric. We multiply each term by 1/4 to get the next term. We start with 8 as the first term. The first five terms are 8, 2, 1/2, 1/8, 1/32.

============================

Problem 3

Similar to problem 1. This time we're adding -19 to each term to get the next one. The starting term is -6. This sequence is arithmetic and the first five terms are -6, -25, -44, -63, -82

============================

Problem 4

Similar to problem 2. This sequence is geometric with common ratio 2/3. First term is 6. The first five terms are 6, 4, 8/3, 16/9, 32/27

5 0
4 years ago
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