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3 years ago

Simplify the expression: –7(5 − u)

Mathematics

2 answers:

Serggg [28]3 years ago

8 0

Answer:

-35+ 7u

Step-by-step explanation:

Distribution Property

distribute the first time by the 2 terms in the parenthesis

-7×5 and -7u

-35 - (-7u)

-35+ 7u

katovenus [111]3 years ago

7 0

Answer:

-7(5-u)

Step-by-step explanation:

-7(5-u)

-7x5 -7x-u

-35+7u

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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

Please help

Murrr4er [49]

<h3>You have the correct answer. It is choice B. Nice work.</h3>

K is the midpoint, so $\overline{IK} \cong \overline{GK}$ and $\overline{JK} \cong \overline{HK}$ , and along with the congruent vertical angles ( $\angle GKH \cong \angle IKJ$ and $\angle GKJ \cong \angle IKH$ ), you would use the SAS (side angle side) congruence theorem to prove the inner pairs of triangles to be congruent.

So, $\triangle HKG \cong \triangle JKI$ (top and bottom triangles) and $\triangle GKJ \cong \triangle IKH$ (left and right triangles).

Then through CPCTC, we can show the corresponding pieces are congruent leading to $\overline{GH} \cong \overline{JI}$ and $\overline{GJ} \cong \overline{HI}$ showing the opposite sides of the quadrilateral are congruent. Therefore we do have a parallelogram and enough information to prove it as such.