PLEASE HELPP PLEASEEE!!!

WILL GIVE BRAINLEIST NEED TO TURN IN BY 9 P.M. PLS HURRY SUPER EASY

yuradex [85]

Answer:

y=2,1

Step-by-step explanation:

If X has to be more than 2 but less than 6 it must be 3,4,5

Because the equation is + that eliminates the answer 5

So the equation can be

3+2=5 or 4+1=5

I hope this helps!

6 0

3 years ago

Find the area of the unshaded reason for these two problems

Taya2010 [7]

9514 1404 393

Answer:

a) 96 m²

b) 26,000 ft²

Step-by-step explanation:

The area of the unshaded region is the difference between the overall area of the rectangle and the area of the shaded region. It can also be found directly using the dimensions of the unshaded region.

a) We have the height of the right triangle, but need to know its base. The Pythagorean theorem can help with that. Let the base of the triangle be represented by x. Then ...

x² +16² = 20²

x² = 400 -256 = 144

x = √144 = 12

The area of the unshaded triangle is given by the formula ...

A = (1/2)bh

A = (1/2)(12)(16) = 96 . . . . square meters

__

b) The shaded area at upper left is a square 100 ft on a side. Its area is ...

A = s² = (100 ft)² = 10,000 ft²

The shaded area at upper right is a triangle with base and height 120 ft and 150 ft. Its area is ...

A = (1/2)bh = (1/2)(120 ft)(150 ft) = 9,000 ft²

The area of the overall rectangle is ...

A = LW = (300 ft)(150 ft) = 45,000 ft²

So, the unshaded area is ...

unshaded area = overall area - left shaded area - right shaded area

unshaded area = 45,000 ft² -10,000 ft² -9,000 ft² = 26,000 ft²

7 0

3 years ago

Evaluate -a - 12a 2 for a = -1

Serhud [2]

The answer is -11
You put "-1" in place of all the "a"s and get
-(-1) - 12(-1)^2
-(-1) is just 1, and (-1)^2 is "1"
We are then left with 1-12(1) which is 1-12 which is -11

6 0

2 years ago

10. 10. If AB bisects CAF, and m<EAF = 72°, then the m<BAF =

never [62]

The measure of ∠BAF is 54°.

Solution:

DF and CE are intersecting lines.

m∠EAF = 72° and AB bisects ∠CAF.

∠EAF and ∠DAC are vertically opposite angles.

Vertical angle theorem:

If two lines are intersecting, then vertically opposite angles are congruent.

∠DAC ≅ ∠EAF

m∠DAC = 72°

Sum of the adjacent angles in a straight line = 180°

m∠DAE + m∠EAF = 180°

m∠DAE + 72° = 180°

Subtract 72° from both sides.

m∠DAE = 108°

∠CAF and ∠DAE are vertically opposite angles.

⇒ m∠CAF = m∠DAE

⇒ m∠CAF = 108°

AB bisects ∠CAF means ∠CAB = ∠BAF

m∠CAB + m∠BAF = 108°

m∠BAF + m∠BAF = 108°

2 m∠BAF = 108°

Divide by 2 on both sides, we get

m∠BAF = 54°

Hence the measure of ∠BAF is 54°.

4 0

3 years ago

Consider the question of whether the home team wins more than half of its games in the National Basketball Association. Suppose

spayn [35]

Answer:

0.0037 = 0.37% probability that the home team would win 65% or more of its games in a simple random sample of 80 games

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$