Please help asap 45 pts

In the diagram below of triangle EFG, H is a midpoint of EF and J is a midpoint

Korolek [52]

Answer:

14

Step-by-step explanation: 2(4x+6)=

\,\,34-3x

34−3x

Since HJ is a midsegment, twice HJ is equal to EG.

8x+12=

\,\,34-3x

34−3x

Distribute.

8x+12=

\,\,-3x+34

−3x+34

Communicative property to change the order

\color{red}{+3x}\phantom{+12}\phantom{=}

+3x+12=

\,\,\color{red}{+3x}\phantom{+34}

+3x+34

+3x to both sides.

11x+12=

\,\,34

34

\phantom{11x}\color{red}{-12}\phantom{=}

11x−12=

\,\,\color{red}{-12}

−12

-12 to both sides.

11x=

\,\,22

22

\frac{11x}{11}=

11

11x

=

\,\,\frac{22}{11}

11

22

Divide both sides by 11

x=

\,\,2

2

Value of x

HJ=

\,\,4x+6

4x+6

Value of HJ

HJ=

\,\,4(2)+6

4(2)+6

Plug in x.

HJ=

\,\,8+6

8+6

Multiply.

HJ=

\,\,14

14

8 0

3 years ago

Please I really need help with this

Delvig [45]

Answer:

Step-by-step explanation:

㏒ 3+㏒(x+2)=1

㏒3(x+2)=1

3(x+2)=10^1

3x+6=10

3x=10-6=4

x=4/3

so A

3.

$log_{2}x+log_{2}(x+2)=3\\or~log_{2}[x(x+2)]=3\\x(x+2)=2^3\\x^2+2x-8=0\\x^2+4x-2x-8=0\\x(x+4)-2(x+4)=0\\(x+4)(x-2)=0\\x=2,-4\\x=-4 ~is~an~extraneous~solution.\\B$

6 0

3 years ago

Given f(x)=1/3x2−1 and g(x)=f(x−5), write an equation that represents g in terms of x.

Naya [18.7K]

Answer:

(fx^2)/3 - 6f

Step-by-step explanation:

8 0

2 years ago

6. they were 300 people at a football match and 35% were adults, the rest were children a. what percentage were children

ioda

Answer: 65%, 195 children

Step-by-step explanation: To get this, we just do 100-35 to get 65 percent. We can calculate this further, and do 300*0.65, and we get 195. Thus, there were 195 children.

4 0

2 years ago

Read 2 more answers

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

5 0

3 years ago