Simplify 8(x-4)+3x-2

What is the slope of OP

frozen [14]

Answer:

B

Step-by-step explanation:

Area of a triangle is 1/2 * Base * height

So we can say: 1/2 * k * (k-2)

That is equal to 24. Let's solve for k first:

$\frac{1}{2}(k)(k-2)=24\\k(k-2)=24*2\\k(k-2)=48$

Putting 8 into k, solves the equation, hence k is 8.

So point P (8, 8- 2) , which is

P(8,6)

The slope of OP is the change in y divided by change in x. Looking at the diagram, going from (0,0) to (8,6), the change in y is 6 and change in x is 8

So slope is 6/8

Reduced, that is 3/4

Answer choice B is right.

8 0

4 years ago

Choose all the expressions that are equal to 0.7 × 0.41.

dlinn [17]

Answer:

None of the expressions is equal to 0.7 x 0.41.

Step-by-step explanation:

Given that 0.7 x 0.41 is equal to 0.287, to determine which options are equal to said operation, the calculations set out for each of them must be performed:

A. 7,100 × 41100 = 291,810,000

B. 710 × 41100 = 29,181,000

C. 7100 × 4110 = 29,181,000

D. 710 × 4110 = 2,918,100

E. 70100 × 41100 = 2,918,100,000

Therefore, looking at all the results produced by the predisposed calculations, it is evident that none of the expressions is equal to 0.7 x 0.41.

7 0

3 years ago

Write the equation for a line that has an initial value of 3 and 3/4 as it’s rate of change

frutty [35]

the slope goes by several names

• average rate of change

• rate of change

• deltaY over deltaX

• Δy over Δx

• rise over run

• gradient

• constant of proportionality

however, is the same cat wearing different costumes.

initial value of 3, namely when x = 0, y = 3, so we have the point (0 , 3) and it has a rate or slope of 3/4.

$(\stackrel{x_1}{0}~,~\stackrel{y_1}{3})\qquad \qquad \stackrel{slope}{m}\implies \cfrac{3}{4} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{\cfrac{3}{4}}(x-\stackrel{x_1}{0})\implies y=\cfrac{3}{4}x+3$

3 0

3 years ago

8. Jenny is getting dressed for school,

UkoKoshka [18]

Answer:

$\huge\boxed{\bf\: Probability = \frac{4}{9}}$

Step-by-step explanation:

According to the given information,

Pairs of black pants = 2
Pair of brown pants = 1
Pairs of blue pants = 2

Then, total pairs of pants = 2 + 1 + 2 = 5

Similarly,

Pink T-shirts = 2
Blue T-shirts = 2

Then, total number of T-shirts = 4

This gives us the total nunber of clothing articles with Jenny, i.e.,

Total pairs of pants + Total number of T-shirts

= 5 + 4

= 9

Now, the probability of pulling out a pair of black pants & a blue T-shirt will be:

Total pairs of black pants + Total nunber of blue T-shirts/Total clothing articles

= 2 + 2 / 9

= 4/9

$\rule{150}{2}$

8 0

3 years ago

A survey conducted by the Consumer Reports National Research Center reported, among other things, that women spend an average of

Nookie1986 [14]

Answer:

(a) The probability that a randomly selected woman shop exactly two hours online is 0.217.

(b) The probability that a randomly selected woman shop 4 or more hours online is 0.0338.

(c) The probability that a randomly selected woman shop less than 5 hours online is 0.9922.

Step-by-step explanation:

Let X = time spent per week shopping online.

It is provided that the random variable X follows a Poisson distribution.

The probability function of a Poisson distribution is:

$P (X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0,1,2,...$

The average time spent per week shopping online is, λ = 1.2.

(a)

Compute the probability that a randomly selected woman shop exactly two hours online over a one-week period as follows:

$P (X=2)=\frac{e^{1.2}(1.2)^{2}}{2!} =0.21686\approx0.217$

Thus, the probability that a randomly selected woman shop exactly two hours online is 0.217.

(b)

Compute the probability that a randomly selected woman shop 4 or more hours online over a one-week period as follows:

P (X ≥ 4) = 1 - P (X < 4)

= 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

$=1-\frac{e^{1.2}(1.2)^{0}}{0!}-\frac{e^{1.2}(1.2)^{1}}{1!}-\frac{e^{1.2}(1.2)^{2}}{2!}-\frac{e^{1.2}(1.2)^{2}}{3!}\\=1-0.3012-0.3614-0.2169-0.0867\\=0.0338$

Thus, the probability that a randomly selected woman shop 4 or more hours online is 0.0338.

(c)

Compute the probability that a randomly selected woman shop less than 5 hours online over a one-week period as follows:

P (X < 5) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

$=\frac{e^{1.2}(1.2)^{0}}{0!}+\frac{e^{1.2}(1.2)^{1}}{1!}+\frac{e^{1.2}(1.2)^{2}}{2!}+\frac{e^{1.2}(1.2)^{3}}{3!}+\frac{e^{1.2}(1.2)^{4}}{4!}\\=0.3012+0.3614+0.2169+0.0867+0.0260\\=0.9922$

Thus, the probability that a randomly selected woman shop less than 5 hours online is 0.9922.

8 0

4 years ago