Find the area of the shaded figure, rounded to two decimals. (Assume x = 3 and y = 4.)

Jessica wants to buy a purse that costs $25.00. She has already saved

jekas [21]

Answer:

$6

Step-by-step explanation:

19-25

4 0

3 years ago

Read 2 more answers

An experiment is completed where a coin is flipped and a six-sided die (singular for dice) is rolled. The outcomes from the expe

victus00 [196]

Answer:

1) $16\frac{2}{3}$ % or 20%

Step-by-step explanation:

Given that:

An experiment which is carried out by flipping a coin and rolling six sided die.

The outcomes can be:

{H1, H2, H3, H5, H6, T1, T1, T2, T3, T5}

As per question statement:

Number of outcomes having tails and an odd number greater than one is 2 (i.e. T3 and T5)

Total number of outcomes here is 10.

Therefore, to find the percentage, we use the following formula:

$\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\times 100$

$\Rightarrow \dfrac{2}{10}\times 100 = 20\%$

As per question statement, the answer is 20%.

If we consider all the outcomes, i.e.

{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

Total number of outcomes = 12

The percentage will come out be :

$\Rightarrow \dfrac{2}{12}\times 100 = 16\frac{2}{3}\%$

7 0

2 years ago

Im so confused with this question?

alexandr1967 [171]

Answer:

1.

Step-by-step explanation:

x intercpet is 3

4/3 is the slope

6 0

3 years ago

Need a little help

nataly862011 [7]

Answer:

500*(106/100)^5=669.11

Step-by-step explanation:

8 0

3 years ago

In a certain city, the daily consumption of electric power, in millions of kilowatt-hours, is a random

kupik [55]

Answer:

a) $\alpha = 3, \beta = 2$

b) 0.0620

Step-by-step explanation:

We are given the following in the question:

Population mean, $\mu$ = 6

Variance, $\sigma^2$ = 12

a) Value of $\alpha, \beta$

We know that

$\alpha \beta = \mu = 6\\\alpha \beta^2 = \sigma^2 = 12$

Dividing the two equations, we get,

$\dfrac{\alpha\beta^2}{\alpha\beta} = \dfrac{12}{6}\\\\\Rightarrow \beta = 2\\\alpha \beta = 6\\\Rightarrow \alpha = 3$

b) probability that on any given day the daily power consumption will exceed 12 million kilowatt hours.

We can write the probability density function as:

$f(x,3,2) = \dfrac{1}{2^{3}(3-1)!}x^{3-1}e^{-\frac{x}{2}}, x > 0\\\\f(x,3,2) = \dfrac{1}{16}x^{2}e^{-\frac{x}{2}}, x > 0$

We have to evaluate:

$P(x >12)\\\\= \dfrac{1}{16}\displaystyle\int^{\infty}_{12}f(x)dx\\\\=\dfrac{1}{16}\bigg[-2x^2e^{-\frac{x}{2}}-2\displaystyle\int xe^{-\frac{x}{2}}dx}\bigg]^{\infty}_{12}\\\\=\dfrac{1}{8}\bigg[x^2e^{-\frac{x}{2}}+4xe^{-\frac{x}{2}}+8e^{-\frac{x}{2}}\bigg]^{\infty}_{12}\\\\=\dfrac{1}{8}\bigg[(\infty)^2e^{-\frac{\infty}{2}}+4(\infty)e^{-\frac{\infty}{2}}+8e^{-\frac{\infty}{2}} -( (12)^2e^{-\frac{12}{2}}+4(12)e^{-\frac{12}{2}}+8e^{-\frac{12}{2}})\bigg]\\\\=0.0620$

0.0620 is the required probability that on any given day the daily power consumption will exceed 12 million kilowatt hours.

3 0

3 years ago