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2 years ago

Plz help will name brainiest and 100 points

Mathematics

2 answers:

Travka [436]2 years ago

5 0

Answer:

Hey bby <3 i would have to say it's E

Step-by-step explanation:

xxTIMURxx [149]2 years ago

4 0

Answer:

Step-by-step explanation:

Multiply 3 2/5 by 15

Change the mixed number to an improper fraction

3 2/5 = (5 * 3 + 2) /5 = 17/5

17/5 * 15

17 * 15/5

17 * 3

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Tony's fish tank is a right rectangular prism 50 \,\text{cm}50cm50, space, c, m long by 25\, \text{cm}25cm25, space, c, m wide.

Svetlanka [38]

The volume of the prism is calculated by multiplying the length, L, width,W, and depth or height, H.
V = L x W x H
Fish tank:
V = 50 cm x 25 cm x 25 cm
= 31250 cm³
Depth of the water in tank (with castle - without castle)
d = 57.5 cm - 48 cm = 9.5 cm
Volume of the caste:
V = 9.5 cm x 50 cm x 25 cm = 11875 cm³

6 0

3 years ago

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$