Help please!!!!!!!!!!!!!!

2 groups of 3 tenths pls answer today I need this answer

SSSSS [86.1K]

Answer:

6/10

Step-by-step explanation:

2 groups of 3/10 =

2 × 3/10 =

6/10

3 0

3 years ago

Help plz! Will mark as brainiest JUST HELP PLZ

Volgvan

1) The lengths of the other 2 sides are 51 feet.

Isosceles triangles have at least 2 equal sides. The base is 60 feet, so we can assume that the other 2 sides are equal

162 = 60 + 2x

x = the equal sides of the triangle.

162 - 60 = 102

102 = 2x

Divided by 2

51 = x

51 + 51 = 102. 102 + 60 = 162

2)

Pool 1: 311.38ft.

8*19.4 = 155.2

8 * 5.7 * 2 = 91.2

5.7 * 5.7 = 32.49/2 (because its a triangle) = 16.254*4 = 64.98

155.2 + 91.2 + 64.98 = 311.39

Pool 2: Either 863.94ft or 484 + 121πft

22 * 22 = 484

22/2 = 11, So, the radius of the half circle is 11

Circle area = (pi)r^2

11*11 = 121

pi * 121 = 121pi

Using 3.14 substitute: 3.14 * 121 = 379.94

379.94 + 484 = 863.94

Pool 3: 207ft

Rhombus area = pq/2

18 * 12 = 216

216/2 = 108

11 * 18 = 198/2 = 99

99 + 108 = 207

7 0

3 years ago

The ratio of boys to girls in Mr. Hakeem's class is 64. If there are 18 boys, how many girls

Bad White [126]

Answer: C. 12 girls

Step-by-step explanation:

6 0

3 years ago

For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains

AnnZ [28]

Answer:

a) There is a 9% probability that a drought lasts exactly 3 intervals.

There is an 85.5% probability that a drought lasts at most 3 intervals.

b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

Step-by-step explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

$f(x) = (1-p)^{x}p$

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

$\mu = \frac{1-p}{p}$

The standard deviation of the geometric distribution is given by the following formula:

$\sigma = \sqrt{\frac{1-p}{p^{2}}$

In this problem, we have that:

$p = 0.383$

So

$\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61$

$\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05$

(a) What is the probability that a drought lasts exactly 3 intervals?

This is $f(3)$

$f(x) = (1-p)^{x}p$

$f(3) = (1-0.383)^{3}*(0.383)$

$f(3) = 0.09$

There is a 9% probability that a drought lasts exactly 3 intervals.

At most 3 intervals?

This is $P = f(0) + f(1) + f(2) + f(3)$

$f(x) = (1-p)^{x}p$

$f(0) = (1-0.383)^{0}*(0.383) = 0.383$

$f(1) = (1-0.383)^{1}*(0.383) = 0.236$

$f(2) = (1-0.383)^{2}*(0.383) = 0.146$

Previously in this exercise, we found that $f(3) = 0.09$

So

$P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855$

There is an 85.5% probability that a drought lasts at most 3 intervals.

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

This is $P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4)$ .

We are working with discrete data, so 3.66 is rounded up to 4.

Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:

$P(X \leq 3) + P(X \geq 4) = 1$

$0.855 + P(X \geq 4) = 1$

$P(X \geq 4) = 0.145$

There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

8 0

3 years ago

I need help with this please

Vitek1552 [10]

The answer to 8a. is 17, the answer to 8b. is 3.9, the answer to 8c. is 1 / 2, the answer to 8d. is 7 7 / 8, the answer to 9a. is commutative property, the answer to 9b. is multiplicative identity property, the answer to 9c. is associative property, and the answer to 9d. is additive inverse property.

Regarding the whole "explain using mental math" thing, I pretty much just used the fact that 8a. and 8b. both had the same number of decimal places. For 8c., I just made each number have a denominator of 10. And last, for 8d., all of the denominators were already the same, which made it pretty easy. I apologize for this section of my answer being so informal lol, I haven't had to do these kinds of problems in like 5 years <span />

7 0

3 years ago