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nlexa [21]
4 years ago
8

Which expression represents a unknown number x is no more than 6?x ≤ 6x < 6x > 6x ≥ 6

Mathematics
1 answer:
aleksley [76]4 years ago
3 0
X is less than or equal to 6
(sorry I am not able to put the symbol on my phone)
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In △BCD, H is the centroid. If HF=6 find BF.
astra-53 [7]

Answer:

18

Step-by-step explanation:

The centroid of a triangle is located ⅓ of the distance from each side to the opposite vertex

So HF=⅓BF

So BF=3xHF=3x6=18

8 0
3 years ago
5x - 6y is less than or equal to 30 in standard form
Marina86 [1]

Answer:

5x - 6y ≤ 30

Step-by-step explanation:

That'd be:  5x - 6y ≤ 30.

7 0
4 years ago
Find the GCF of each set of numbers <br><br> 150, 200, and 300
jolli1 [7]
Your GCF is 50 of each set of numbers
3 0
3 years ago
Multiply.<br> 1 2/9X 1 4/5
alina1380 [7]

Answer:

Equals 2.2

Step-by-step explanation:

hope it's useful

6 0
3 years ago
Read 2 more answers
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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