All categories

3 years ago

PLEASE ANSWER THIS QUESTION ILL MARK YOU AS THE BRAINLIEST

Mathematics

1 answer:

Mrac [35]3 years ago

8 0

4. (x+6)^2=51

This is because when you multiply x+6 by itself you get
x^2+12x+36=51

then you need to move the 51 to the other side

x^2+12x-15=0

You might be interested in

What is the product ? (7x^2 )(2x^3 +5)(x^2-4x-9)

eimsori [14]

Answer:

Step-by-step explanation:

5 0

4 years ago

What is the answer to 6w^2-24w-126

Dafna1 [17]

Answer:

6(w+3)(w-7)

Sorry if this is the wrong answer. But I hope you get a good grade on your test.

7 0

4 years ago

A 15-foot ladder must make an angle of 30° with the ground if it is to reach a certain window. What angle must a 20-foot ladder

Tju [1.3M]

The answer is A) 22. Not 48.

3 0

4 years ago

PLEASE HELP IT’S DUE TMR

Soloha48 [4]

4 + 3 + 1 = 8

450000 divided by 8 = 56250

The first person gets 225000.

The second person gets 168750.

The last person gets 56250.

pls give me brainliest if it helped:)

4 0

3 years ago

A national college researcher reported that 65% of students who graduated from high school in 2012 enrolled in college. Twenty n

Usimov [2.4K]

Answer:

a) The probability that exactly 17 of them enroll in college is 0.116.

b) The probability that more than 14 enroll in college is 0.995.

c) The probability that fewer than 11 enroll in college is 0.001.

d) It would be be unusual if more than 24 of them enroll in college since the probability is 0.009.

Step-by-step explanation:

We can model this with a binomial distribution, with n=29 and p=0.65.

The probability that k students from the sample who graduated from high school in 2012 enrolled in college is:

$P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{29}{k} 0.65^{k} 0.35^{29-k}\\\\\\$

a) The probability that exactly 17 of them enroll in college is:

$P(x=17) = \dbinom{29}{17} p^{17}(1-p)^{12}=51895935*0.0007*0=0.116\\\\\\$

b) The probability that more than 14 of them enroll in college is:

$P(X>14)=\sum_{15}^{29} P(X=k_i)=1-\sum_{0}^{14} P(X=k_i)\\\\\\P(x=0)=0\\\\P(x=1)=0\\\\P(x=2)=0\\\\P(x=3)=0\\\\P(x=4)=0\\\\P(x=5)=0\\\\P(x=6)=0\\\\P(x=7)=0\\\\P(x=8)=0\\\\P(x=9)=0\\\\P(x=10)=0.001\\\\P(x=11)=0.002\\\\P(x=12)=0.005\\\\P(x=13)=0.013\\\\P(x=14)=0.027\\\\\\P(X>14)=1-0.005=0.995$

c) Using the probabilities calculated in the point b, we have:

$P(X$

d) The probabilities that more than 24 enroll in college is:

$P(X>24)=\sum_{25}^{29}P(X=k_i)\\\\\\ P(x=25) = \dbinom{29}{25} p^{25}(1-p)^{4}=23751*0*0.015=0.007\\\\\\P(x=26) = \dbinom{29}{26} p^{26}(1-p)^{3}=3654*0*0.043=0.002\\\\\\P(x=27) = \dbinom{29}{27} p^{27}(1-p)^{2}=406*0*0.123=0\\\\\\P(x=28) = \dbinom{29}{28} p^{28}(1-p)^{1}=29*0*0.35=0\\\\\\P(x=29) = \dbinom{29}{29} p^{29}(1-p)^{0}=1*0*1=0\\\\\\\\P(X>24)=0.007+0.002+0+0+0=0.009$

6 0

3 years ago