Give me a fraction that is greater than 1/3

An equilateral triangle ΔLMQ (see picture to the right) is drawn in the interior of square LMNP so that side is common. Find m∠L

lesya692 [45]

I don’t see the picture if I see it I can help you tho

5 0

3 years ago

During a musical, an orchestra is playing. As the music plays, the volume changes at the beginning of the piece can be modeled b

Ivanshal [37]

Answer:

what

Step-by-step explanation:

...

5 0

3 years ago

Read 2 more answers

A number Q has a remainder of 3 when divided by 5 and also has a remainder of 4 when divided by 8. What is the smallest value Q

coldgirl [10]

5555666689919374747373

6 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

In Exercises 15-18, find the measure of the exterior angle.

valina [46]

Answer:

The measure of the exterior angle is 92 degrees

Step-by-step explanation:

Problem 16)

we know that

An exterior angle of a triangle is equal to the sum of the opposite interior angles

so

In this problem

$(2x-2)\°=x\°+45\°$

solve for x

$2x-x=45+2$

$x=47\°$

Find the measure of the exterior angle

$(2x-2)\°=(2(47)-2)=92\°$

6 0

3 years ago