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3 years ago

I will give you Brainiest if you are right.

Mathematics

2 answers:

Vaselesa [24]3 years ago

7 0

B is the correct answer

vredina [299]3 years ago

4 0

It’s 8x+12 — just use the distributive property : a(b+c) = to ab + ac!

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Can you help plz!!!!

Zolol [24]

Okay so the table represents B, in one hour of work, the person will make 18.25 (you get this by dividing 36.50 by two because that is how much they made in two hours, we need to know how much they make in one)
Therefore, the equation for A, which has to be less than B, would be either be Y=18x or Y=18.1x

8 0

3 years ago

At 1 P.M., the total snowfall is 5 centimeters. At 4 P.M., the total snowfall is 16 centimeters. What is the mean hourly snowfal

Yuri [45]

Answer:

Mr. Snowbully

Step-by-step explanation:

I think the mean hourly snowfall is named Mr. Snowbully

It's a funny name!

Well, the snow would grow at a rate of 4 cm every hour so...

5 0

3 years ago

Mr. Lopez and Ms. Spence both went on separate road

ziro4ka [17]

Answer:

950 miles

the problem states that Mr. Lopez drove 77 more miles so just add 873+77 to get 950 miles

7 0

3 years ago

Your flight has been delayed: At Denver International Airport, 85% of recent flights have arrived on time. A sample of 14 flight

Bezzdna [24]

Answer:

a. p=1.000

b. p=0.2924

c. p=0.7358

d. No

Step-by-step explanation:

a. This problem satisfies all the criteria for a binomial experiment expressed as:

$P(X=x){n\choose x}p^x(1-p)^{n-x}$

-Given that p=0.85, n=14, the probability that exactly all 14 were on time is calculated as:

$P(X\geq 1)=1-P(X=0)\\\\=1-{12\choose 0}0.85^0(1-0.85)^{12}\\\\=1-1.297\times 10^{-10}\\\\=1.0000$

Hence, the probability that all 12 flights are on time is 1.0000

b. Given that n=12, and p=0.85

-The probability that exactly 10 flights are on time is calculated as;

$P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\={12\choose 10}0.85^{10}(1-0.85)^{2}\\\\=0.2924$

Hence, the probability that exactly 10 flights are on time is 0.2924

c. Given that n=12, and p=0.85

-The probability that more of 10 or more flights are on time:

$P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq 10)=P(X=10)+P(X=11)+P(X=12)\\\\={12\choose 10}0.85^{10}(0.15)^{2}+{12\choose 11}0.85^{11}(0.15)^{1}+{12\choose 12}0.85^{12}(0.15)^{0}\\\\=0.2924+0.3012+0.1422\\\\=0.7358$

Hence, the probability of 10+ flights being on time is 0.7358

d. We first find the mean of the distribution:

$\mu=E(X)=0.85\times 14\\\\=11.9$

#We then find the probability of 11+=0.3012+0.1422=0.4434

-We compare the expectation to the probability of 11+ flights being on time.

No. Since the probability $P(X\geq 10)$ =0.4434 < that the expectation, 11.9, it is not unusual for 11+ flights to be on time.

*I have used a sample size of n=12 since there are two separate n values:

5 0

3 years ago

Solve for x <br> 42=x-2(x+4)

aniked [119]

Answer:

-50

Step-by-step explanation:

42+=x-2x-8

42=-x-8

(add 8 to both sides)

50=-x

(divide by negative 1)

-50=x

7 0

2 years ago