F (7-3)2 x 7x+4 = 77, find the value of x.

Which congruence theorems can be used to prove ΔEFG ≅ ΔJHG? Select two options.

mezya [45]

<h2>Answer: </h2><h2> </h2>

ASA and AAS

<h2>Step-by-step explanation: </h2>

We do not know if these are right triangles; therefore we cannot use HL to prove congruence.

We do not have 2 or 3 sides marked congruent; therefore we cannot use SSS or SAS to prove congruence.

We are given that EF is parallel to HJ. This makes EJ a transversal. This also means that ∠HJG and ∠GEF are alternate interior angles and are therefore congruent. We also know that ∠EGF and ∠HGJ are vertical angles and are congruent. This gives us two angles and a non-included side, which is the AAS congruence theorem.

Since EF and HJ are parallel and EJ is a transversal, ∠JHG and ∠EFG are alternate interior angles and are congruent. Again we have that ∠EGF and ∠HGJ are vertical angles and are congruent; this gives us two angles and an included side, which is the ASA congruence theorem.

5 0

2 years ago

Read 2 more answers

John can type 2/3 of a manuscript in 12 hours. How much of the manuscript can John write in 1 hour?

Marina86 [1]

John can write 1/18 of the manuscript in 1 hour
Hope this helps!!

8 0

2 years ago

Read 2 more answers

Shelly bought a house five years ago for $150,000 and obtained an 80% loan. Now the home is worth $140,000 and her loan balance

LiRa [457]

Answer:

The current equity is $32,000

Step-by-step explanation:

Here, we are to calculate the current equity of the House

Value of house initially is $150,000

loan borrowed is 80% of this

= 80/100 * 150,000 = $120,000

Now the house value is 140,000 and the loan amount is reduced to 12,000

The outstanding loan amount to pay would be 120,000-12,000 = 108,000

The current equity is thus = Assets - liability = 140,000-108,000 = $32,000

5 0

2 years ago

Jack was so frustrated with his slow laptop that he threw it out of his second story window. The height, h, of the laptop at tim

jeyben [28]

Answer:

The domain of the function is the interval [0,2.23]

see the explanation

Step-by-step explanation:

Let

t ----> the time in seconds

h(t) ----> the height of the laptop in units

we have

$h(t)=-16t^{2}+28t+17$

we know that

When the laptop hits the ground, the value of h(t) is equal to zero

so

For h(t)=0

$-16t^{2}+28t+17=0$

Solve the quadratic equation

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-16t^{2}+28t+17=0$

so

$a=-16\\b=28\\c=17$

substitute in the formula

$x=\frac{-28(+/-)\sqrt{28^{2}-4(-16)(17)}} {2(-16)}$

$x=\frac{-28(+/-)\sqrt{1,872}} {-32}$

$x=\frac{-28(+/-)12\sqrt{13}} {-32}$

$x_1=\frac{-28(+)12\sqrt{13}} {-32}=-0.477$

$x_1=\frac{28(-)12\sqrt{13}} {32}=-0.477$ ---> is not a solution

$x_2=\frac{-28(-)12\sqrt{13}} {-32}$

$x_2=\frac{28(+)12\sqrt{13}} {32}=2.23\ sec$

therefore

The domain of the function is the interval [0,2.23]

All real numbers greater than or equal to 0 seconds and less than or equal to 2.23 seconds

$0\ sec \leq x \leq 2.23\ sec$

5 0

2 years ago

The Wall Street Journal reports that 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deduction

Len [333]

Answer:

(a) n : 20 50 100 500

P (-200 < X - μ < 200) : 0.2886 0.4444 0.5954 0.9376

(b) The correct option is (b).

Step-by-step explanation:

Let the random variable X represent the amount of deductions for taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return.

The mean amount of deductions is, μ = $16,642 and standard deviation is, σ = $2,400.

Assuming that the random variable X follows a normal distribution.

(a)

Compute the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean as follows:

For a sample size of n = 20

$P(\mu-200$

$=P(-0.37$

For a sample size of n = 50

$P(\mu-200$

$=P(-0.59$

For a sample size of n = 100

$P(\mu-200$

$=P(-0.83$

For a sample size of n = 500

$P(\mu-200$

$=P(-1.86$

n : 20 50 100 500

P (-200 < X - μ < 200) : 0.2886 0.4444 0.5954 0.9376

(b)

The law of large numbers, in probability concept, states that as we increase the sample size, the mean of the sample ( $\bar x$ ) approaches the whole population mean ( $\mu_{x}$ ).

Consider the probabilities computed in part (a).

As the sample size increases from 20 to 500 the probability that the sample mean is within $200 of the population mean gets closer to 1.

So, a larger sample increases the probability that the sample mean will be within a specified distance of the population mean.

Thus, the correct option is (b).

8 0

2 years ago