What does 2/5 of 4/5=

Options:<br> 14<br> 36<br> 16<br> 12

Lerok [7]

Answer:

12 million dollars.

Step-by-step explanation:

Let the sales in the first year be x million .

So the sales in the second year were x + 4 million, in the third year were 3(x + 4) and this is 48 million dollars.

So we have the equation 3(x + 4) = 48

3x + 12 = 48

3x = 36

x = 12 million dollars.

8 0

3 years ago

Which equations represent the asymptotes of the hyperbola?

makvit [3.9K]

Answer:

see below

Step-by-step explanation:

The equation of the hyperbola can be written as ...

((x -h)/a)² -((y -k)/b)² = 1

This has asymptotes ...

(x -h)/a ± (y -k)/b = 0

Solving for y, we have ...

y = ±(b/a)(x -h) +k

Filling in the given values a=6, b=8, h=1, k=2, we have ...

y = ±8/6(x -1) +2

$y=\dfrac{\pm4x\mp4+6}{3}\\\\\boxed{y=\dfrac{4x+2}{3}\ \text{and }y=\dfrac{10-4x}{3}}$

4 0

3 years ago

Math help! 20 Points!

erastova [34]

The anwer is 3 because it 3

4 0

3 years ago

Read 2 more answers

A box contains four 10 nf and eight 100 nf capacitors. (a) what is the probability of obtaining at least two 10 nf capacitors if

grigory [225]

there are 4 of the 10-nf and 8 of the 100-nf capacitors which is a total of 12 items.

the probability of drawing a 10-nf is: 4 out of 12 = $\frac{1}{3}$

the probability of drawing a 100-nf is: 8 out of 12 = $\frac{2}{3}$

(a) "at least two" means: 2 or 3 or 4

Probability of 2 (10's) and 2 (100's): $\frac{1}{3}$ x $\frac{1}{3}$ x $\frac{2}{3}$ x $\frac{2}{3}$ = $\frac{4}{81}$

Probability of 3 (10's) and 1 (100's): $\frac{1}{3}$ x $\frac{1}{3}$ x $\frac{1}{3}$ x $\frac{2}{3}$ = $\frac{2}{81}$

Probability of 4 (10's) and 0 (100's): $\frac{1}{3}$ x $\frac{1}{3}$ x $\frac{1}{3}$ x $\frac{1}{3}$ = $\frac{1}{81}$

2 or 3 or 4: $\frac{4}{81}$ + $\frac{2}{81}$ + $\frac{1}{81}$ = $\frac{7}{81}$

(b) without replacement

Probability of 2 (10's) and 2 (100's): $\frac{4}{12}$ x $\frac{3}{11}$ x $\frac{8}{10}$ x $\frac{7}{9}$ = $\frac{4 x 3 x 8 x 7}{12 x 11 x 10 x 9}$

Probability of 3 (10's) and 1 (100's): $\frac{4}{12}$ x $\frac{3}{11}$ x $\frac{2}{10}$ x $\frac{8}{9}$ = $\frac{4 x 3 x 2 x 8}{12 x 11 x 10 x 9}$

Probability of 4 (10's) and 0 (100's): $\frac{4}{12}$ x $\frac{3}{11}$ x $\frac{2}{10}$ x $\frac{1}{9}$ = $\frac{4 x 3 x 2 x 1}{12 x 11 x 10 x 9}$

2 or 3 or 4: $\frac{4 x 3 x 8 x 7}{12 x 11 x 10 x 9}$ + $\frac{4 x 3 x 2 x 8}{12 x 11 x 10 x 9}$ + $\frac{4 x 3 x 2 x 1}{12 x 11 x 10 x 9}$ = $\frac{672 + 192 + 24}{12 x 11 x 10 x 9}$ = $\frac{888}{12 x 11 x 10 x 9}$ = $\frac{111}{1485}$

8 0

3 years ago

Read 2 more answers

Small sample: During an economic downturn, companies were sampled and asked whether they were planning to increase their workfor

leva [86]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The confidence level interval is $0.016 \le C \le 0.404$

Step-by-step explanation:

The sample size is $n = 20$

The number planning to increase workforce is $x = 3$

The confidence level is $c = 98$ %

The value of proportion for a plus 4 method is

$p = \frac{x+2}{n+4}$

substituting values

$p = \frac{3+2}{20+4}$

$p =0.21$

The z-critical value at confidence level of 98% is

$z_{c}=z_{0.98} = 2.33$

This values is obtained from the standard normal table

The confidence level interval can be mathematically represented as

$C =p \ \pm z_{c} * \sqrt{\frac{p(1-p)}{n+4} }$

substituting values

$C = 0.21 \pm 2.33 * \sqrt{\frac{0.21(1- 0.21)}{20 +4} }$

$C = 0.21 \pm 0.194$

=> $0.016 \le C \le 0.404$

7 0

3 years ago