I NEED HELP WITH BOTH I WILL MARK THE BRAINIEST

Help!!!! I will give you brainiest!!!! Pls find the solution of y=3x+5 for x=-2

Sergeeva-Olga [200]

Answer:

I think the answer is B

can you also please mark me as brainliest

5 0

3 years ago

Read 2 more answers

Principal, $3000; Annual interest rate, 5.4%; time, 4 years The Balance after 4 years is?

Neporo4naja [7]

Answer: 777600

Step-by-step explanation:

you times 12 by 4 then you times 3000 by 5.4 and add the two answers

3 0

3 years ago

Which of the following is equal to the rational expression when x -2 or -1? x^2-4/(x+2)(x+1)

Orlov [11]

Answer:

(x+2)(x+1)

Step-by-step explanation:

Assuming im interpreting this correctly, it should be (x+2)(x+1), since if you plug in -2 and -1, it will give you 0, which is the answer.

8 0

3 years ago

We are throwing darts on a disk-shaped board of radius 5. We assume that the proposition of the dart is a uniformly chosen point

Vlad1618 [11]

Answer:

the probability that we hit the bullseye at least 100 times is 0.0113

Step-by-step explanation:

Given the data in the question;

Binomial distribution

We find the probability of hitting the dart on the disk

⇒ Area of small disk / Area of bigger disk

⇒ πR₁² / πR₂²

given that; disk-shaped board of radius R² = 5, disk-shaped bullseye with radius R₁ = 1

so we substitute

⇒ π(1)² / π(5)² = π/π25 = 1/25 = 0.04

Since we have to hit the disk 2000 times, we represent the number of times the smaller disk ( BULLSEYE ) will be hit by X.

so

X ~ Bin( 2000, 0.04 )

n = 2000

p = 0.04

np = 2000 × 0.04 = 80

Using central limit theorem;

X ~ N( np, np( 1 - p ) )

we substitute

X ~ N( 80, 80( 1 - 0.04 ) )

X ~ N( 80, 80( 0.96 ) )

X ~ N( 80, 76.8 )

So, the probability that we hit the bullseye at least 100 times, P( X ≥ 100 ) will be;

we covert to standard normal variable

⇒ P( X ≥ $\frac{100-80}{\sqrt{76.8} }$ )

⇒ P( X ≥ 2.28217 )

From standard normal distribution table

P( X ≥ 2.28217 ) = 0.0113

Therefore, the probability that we hit the bullseye at least 100 times is 0.0113

3 0

3 years ago

Breyers is a major producer of ice cream and would like to test if the average American consumes more than 17 ounces of ice crea

asambeis [7]

Answer:

The conclusion for this hypothesis test would be that the average American consumes less than or equal to 17 ounces of ice cream per month.

Step-by-step explanation:

We are given that Breyers is a major producer of ice cream and would like to test if the average American consumes more than 17 ounces of ice cream per month.

A random sample of 25 Americans was found to consume an average of 19 ounces of ice cream last month. The standard deviation for this sample was 5 ounces.

Let $\mu$ = average ounces of ice cream consumed by American per month

So, Null Hypothesis, $H_0$ : $\mu \leq$ 17 ounces {means that the average American consumes less than or equal to 17 ounces of ice cream per month}

Alternate Hypothesis, $H_A$ : $\mu$ > 17 ounces {means that the average American consumes more than 17 ounces of ice cream per month}

The test statistics that will be used here is One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average = 19 ounces

s = sample standard deviation = 5 ounces

n = sample of Americans = 25

So, test statistics = $\frac{19-17}{\frac{5}{\sqrt{25} } }$ ~ $t_2_4$

= 2

The value of the test statistics is 2.

Now at 0.025 significance level, the t table gives critical value of 2.06 at 24 degree of freedom for right-tailed test. Since our test statistics is less than the critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the the average American consumes less than or equal to 17 ounces of ice cream per month.

8 0

3 years ago