Can someone help me with #3? Thanks!

Part A The measure of ∠AFC is 120°, and the measure of ∠BFC is 95°. Which equation can you use to find the measure of ∠AFB? 120°

aev [14]

Answer:

x = 120-95

Step-by-step explanation:

The correct equation for the expression is:

∠AFC = ∠AFB+ ∠BFC

Given that

∠AFC = 120°

∠BFC = 95°

Substitute

120= ∠AFB+95

∠AFB = 120-95

If ∠AFB = x

The required equation will be:

x = 120-95

3 0

3 years ago

Please help me answer this

Alja [10]

Answer:

Brand A

Step-by-step explanation: Brand A because y=50x+100 and that equals 150x. For brand B y=25x+200 and that equals 225x. So Brand A is lower.

3 0

3 years ago

815.755 round to the nearest hundreath815.755 round to the nearest hundredth

777dan777 [17]

815.755 round to the nearest hundred is 816

4 0

4 years ago

Read 2 more answers

45 points!!! Will mark BRAINLIEST must show explanation!

andrew-mc [135]

Answer:

4hx - 6h

Step-by-step explanation:

Given

f(x) = x^2 - 3x

Simply

f(x + h) - f(x - h)

Solution

f(x + h) - f(x - h) =
(x + h)^2 - 3(x + h) - (x - h)^2 + 3(x - h) =
x^2 + 2hx + h^2 - 3x - 3h - x^2 + 2hx - h^2 + 3x - 3h =
(x^2 - x^2) + (2hx + 2hx) - (3x - 3x) + (h^2 - h^2) - (3h + 3h) =
4hx - 6h

6 0

3 years ago

Recent homebuyers from a local developer allege that 30% of the houses this developer constructs have some major defect that wil

umka21 [38]

Answer:

3.54% probability of observing at most two defective homes out of a random sample of 20

Step-by-step explanation:

For each house that this developer constructs, there are only two possible outcomes. Either there are some major defect that will require substantial repairs, or there is not. The probability of a house having some major defect that will require substantial repairs is independent of other houses. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

30% of the houses this developer constructs have some major defect that will require substantial repairs.

This means that $p = 0.3$

If the allegation is correct, what is the probability of observing at most two defective homes out of a random sample of 20

This is $P(X \leq 2)$ when n = 20. So

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C_{20,0}.(0.3)^{0}.(0.7)^{20} = 0.0008$

$P(X = 1) = C_{20,1}.(0.3)^{1}.(0.7)^{19} = 0.0068$

$P(X = 2) = C_{20,2}.(0.3)^{2}.(0.7)^{18} = 0.0278$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0008 + 0.0068 + 0.0278 = 0.0354$

3.54% probability of observing at most two defective homes out of a random sample of 20

5 0

3 years ago