What is the equation of a line that has a slope of -2 and goes through the point (-9,1)

Christina describes her shape. She says it has 3 equal sides that are each 4 centimeters in length. It has no right angles. Do u

Salsk061 [2.6K]

draw an equilateral triangle that has a side length of all 4cm

4 0

3 years ago

Read 2 more answers

1. What is the value of h? (1 point)

irinina [24]

The value of h is 11.3.

Solution:

Given data:

The triangle is a right triangle.

θ = 45° and opposite side of θ = 8

The value of sin 45° = $\frac{1}{\sqrt{2} }$

Using trigonometric formula,

$$\sin\theta=\frac{\text{Opposite side}}{\text{Hypotenuse}}$

$$\sin45^\circ=\frac{{8}}{h}$

$$\frac{1}{\sqrt{2} } =\frac{{8}}{h}$

Do cross multiplication.

$h = 8\sqrt 2$

h = 11.3

The value of h is 11.3.

7 0

3 years ago

There is a mound of g pounds of gravel in a quarry. Throughout the day, 300 pounds of gravel are added to the mound. Two orders

galina1969 [7]

Answer:

$g+300-700-700=1500, g=2600$

Step-by-step explanation:

At the start of the day there are g pounds of gravel in the quarry.

Then it says that there were 300 pounds added to the mound, so at this point the pounds of gravel in the mound are now g+300.

Then two orders of 700 pounds are sold and removed from the mound. That means they removed 700 pounds from the mound 2 times. Therefore at this point the pounds of gravel in the mound are now g+300-700-700.

Then it says that at this point, the amount of pounds in the mound are 1500. Hence we get

$g+300-700-700=1500$

To solve for g we just add 700 two times to both sides of the equation, and subtract 300 from both sides of the equation, getting:

$g+300-700-700+700+700-300=1500+700+700-300$

And so $g=2600$

4 0

3 years ago

The screening process for detecting a rare disease is not perfect. Researchers have developed a blood test that is considered fa

lesya692 [45]

Answer:

Type I: 1.9%, Type II: 1.6%

Step-by-step explanation:

given null hypothesis

H0=the individual has not taken steroids.

type 1 error-falsely rejecting the null hypothesis

⇒ actually the null hypothesis is true⇒the individual has not taken steroids.

but we rejected it ⇒our prediction is the individual has taken steroids.

typr II error- not rejecting null hypothesis when it has to be rejected

⇒actually null hypothesis is false ⇒the individual has taken steroids.

but we didnt reject⇒the individual has not taken steroids.

let us denote

the individual has taken steroids by 1

the individual has not taken steroids.by 0

predicted

1 0

actual 1 98.4% 1.6%

0 1.9% 98.1%

so for type 1 error

actual-0

predicted-1

therefore from above table we can see that probability of Type I error is 1.9%=0.019

so for type II error

actual-1

predicted-0

therefore from above table we can see that probability of Type I error is 1.6%=0.016

5 0

3 years ago

The computers of nine engineers at a certain company are to be replaced. Four of the engineers have selected laptops and the oth

Gala2k [10]

Answer:

(a) There are 70 different ways set up 4 computers out of 8.

(b) The probability that exactly three of the selected computers are desktops is 0.305.

(c) The probability that at least three of the selected computers are desktops is 0.401.

Step-by-step explanation:

Of the 9 new computers 4 are laptops and 5 are desktop.

Let X = a laptop is selected and Y = a desktop is selected.

The probability of selecting a laptop is = $P(Laptop) = p_{X} = \frac{4}{9}$

The probability of selecting a desktop is = $P(Desktop) = p_{Y} = \frac{5}{9}$

Then both X and Y follows Binomial distribution.

$X\sim Bin(9, \frac{4}{9})\\ Y\sim Bin(9, \frac{5}{9})$

The probability function of a binomial distribution is:

$P(U=k)={n\choose k}\times(p)^{k}\times (1-p)^{n-k}$

(a)

Combination is used to determine the number of ways to select <em>k</em> objects from <em>n</em> distinct objects without replacement.

It is denotes as: ${n\choose k}=\frac{n!}{k!(n-k)!}$

In this case 4 computers are to selected of 8 to be set up. Since there cannot be replacement, i.e. we cannot set up one computer twice or thrice, use combinations to determine the number of ways to set up 4 computers of 8.

The number of ways to set up 4 computers of 8 is:

${8\choose 4}=\frac{8!}{4!(8-4)!}\\=\frac{8!}{4!\times 4!} \\=70$

Thus, there are 70 different ways set up 4 computers out of 8.

(b)

It is provided that 4 computers are randomly selected.

Compute the probability that exactly 3 of the 4 computers selected are desktops as follows:

$P(Y=3)={4\choose 3}\times(\frac{5}{9})^{3}\times (1-\frac{5}{9})^{4-3}\\=4\times\frac{125}{729}\times\frac{4}{9}\\ =0.304832\\\approx0.305$

Thus, the probability that exactly three of the selected computers are desktops is 0.305.

(c)

Compute the probability that of the 4 computers selected at least 3 are desktops as follows:

$P(Y\geq 3)=1-P(Y$

Thus, the probability that at least three of the selected computers are desktops is 0.401.

6 0

3 years ago