Help please this is hard

Serga [27]

36/48=6/8=3/4 so 75% completed training

8 0

3 years ago

A rectangular countertop has a perimeter of 26 feet. its area is 30 square feet. what are the dimensions of the counter?

Sladkaya [172]

P=26=2*(l+L)
⇒ l+L=26:2=13 feet
l=3, L=10

A=l*L=30=10*3

⇒ l=3 feet
L=10 feet

4 0

3 years ago

A class of 40 students elected a class president. There were 12 votes for Candidate A, and 18 votes for Candidate B.

Helen [10]

Hey there, Bulgogi!

The correct answer will be the first one since [I'm assuming] you know that to calculate the percentage of something, it's the amount of that something divided by the total amount but since only 30 students voted [12+18] multiplied by 100.

So, we are calculating the percentage on votes and 12 students voted for Candidate A so our Numerator will be 12 and our Denominator will be the total amount of student who voted, so 30 [or 12+18]

Once put into a fraction, we will get $\frac{12}{12+18}$ which would be equals to 2/5 or 0.4. Now we multiply by 100 and we get a total of 40% for Candidate A.

Thank you for using Brainly.
See you soon!

4 0

4 years ago

Calculate the value of 1/3 a = -5

sladkih [1.3K]

Answer:

- 15

Step-by-step explanation:

$\frac{1}{3} a = - 5 \\ \\ a = - 5 \times 3 \\ \\ a = - 15$

6 0

2 years ago

Read 2 more answers

The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(

Umnica [9.8K]

Answer:

(a) $A_1$ and $A_2$ are indeed mutually-exclusive.

(b) $\displaystyle P(A_1\; \cap \; B) = \frac{1}{20}$ , whereas $\displaystyle P(A_2\; \cap \; B) = \frac{1}{25}$ .

(c) $\displaystyle P(B) = \frac{9}{100}$ .

(d) $\displaystyle P(A_1 \; |\; B) \approx \frac{5}{9}$ , whereas $P(A_1 \; |\; B) = \displaystyle \frac{4}{9}$

Step-by-step explanation:

$P(A_1 \; \cap \; A_2) = 0$ means that it is impossible for events $A_1$ and $A_2$ to happen at the same time. Therefore, event $A_1$ and $A_2$ are mutually-exclusive.

By the definition of conditional probability:

$\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}$ .

Rearrange to obtain:

$\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot P(A_1) = 0.25 \times 0.20 = \frac{1}{20}$ .

Similarly:

$\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot P(A_2) = 0.80 \times 0.05 = \frac{1}{25}$ .

Note that:

$\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}$ .

In other words, $A_1$ and $A_2$ are collectively-exhaustive. Since $A_1$ and $A_2$ are collectively-exhaustive and mutually-exclusive at the same time:

$\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}$ .

By Bayes' Theorem:

$\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}$ .

Similarly:

$\begin{aligned} P(A_2 \; |\; B) &= \frac{P(B \; | \; A_2) \cdot P(A_2)}{P(B)} \\ &= \frac{0.05 \times 0.80}{9/100} = \frac{0.04 \times 100}{9} = \frac{4}{9}\end{aligned}$ .

6 0

3 years ago