Solve for x, if 2x-5=7

Given that m∠ABD = 84° and m∠BAC = 37°, what is the m∠ACD?

stich3 [128]

Answer: 84°

Step-by-step explanation:

Pardon my drawing, I know it;s not prefect, and I'm not sure how to draw a figure on Brainly, but this is the best I can do. For this problem, it is easier to draw it out so you can see that the angles look like. Just pretend the lines are filled in, and this is a perfect parallelogram.

A ________________ C

/ /

B /_______________/ D

As you can see, the angles given in the problem gives us a parallelogram. Since we know m∠ABD is 84° and m∠BAC is 37°, we know that m∠ACD is 84°. If you draw diagonal lines, ∠ABD and ∠ACD are vertical angles. Vertical angles are congruent. Therefore, m∠ACD is 84°.

7 0

2 years ago

Brainliest available to the correct answer!

Andrew [12]

What is the question? :)

8 0

2 years ago

What is 7 grams in milligrams?

hodyreva [135]

Answer:

7000 milligrams. hope this help

3 0

2 years ago

Jane is is charge of making a banner for the basketball game this Saturday. She measures how long the banner is before painting

RSB [31]

Answer:

The measurement which Jane finds to be 10 meters is the length of the banner.

Step-by-step explanation:

The measurement of 10 meters which Jane found after measuring how long the banner is before painting is the LENGTH of the banner.

This is clear from the unit of what she finds (meters). It only indicates the measurement of one part of the banner, even though a banner has two parts, the length and width.

It is possible to find the AREA, or PERIMETER, or LENGTH.

But what she finds is the LENGTH of the banner. If it was Area or Perimeter, the unit would have been square meters.

7 0

2 years ago

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

2 years ago