Please help 30 points and brainly answer :(

Yolanda needs 5 pounds of ground beef to make lasagna for a family reunion. One package of ground beef holds 2 1/2 pounds. Anoth

jonny [76]

You can find the answer by adding the two packages and subtracting them by 5.
2 1/2+2 3/5=5 1/10
5 1/10-5=1/10

Yolonda will have 1/10 left over.

4 0

2 years ago

A circle has a diameter of 9cm

Katen [24]

For the square to fit inside the circle without touching it, the diagonal of the square needs to be less than the diameter of the circle.

Using the Pythagorean theorem we can calculate the diagonal of the square:

X = SQRT(5^2 + 5^2)

X = SQRT(25 + 25)

X = SQRT(50)

X = 7.07

The diagonal of the square is 7.07 cm, which is less than the diameter of the circle, 9cm, so it will fit .

8 0

3 years ago

Lucas is making accessories for the soccer team. He uses 883.68 inches of fabric on headbands for 46 players and 2 coaches. He a

zimovet [89]

The total fabric used for headband and wristband of a player was: 27.84 inches

Step-by-step explanation:

The question will be solved step by step.

So,

Total ribbon used for headbands = 883.68 inches

Total number of people for whom headbands were made = 46 players + 2 coaches = 48 inches

so the headband for one person will use fabric: $\frac{883.68}{48}\\= 18.41\ inches$

Then

Total fabric used for wristbands = 433.78 inches

Total players = 46

$Fabri\ for\ one\ player's\ wristband = \frac{433.78}{46}\\=9.43\ inches$

Total fabric for one player = 18.41 + 9.43 = 27.84 inches

Hence,

The total fabric used for headband and wristband of a player was: 27.84 inches

Keywords: Fractions, decimals

Learn more about fractions at:

#LearnwithBrainly

6 0

2 years ago

What is the midpoint of BC

torisob [31]

Step-by-step explanation:

midpoint of BC is 63 middle of 60 and 65

5 0

2 years ago

Let X be a random variable with probability mass function P(X = 1) = 1 2 , P(X = 2) = 1 3 , P(X = 5) = 1 6 (a) Find a function g

Goryan [66]

The question is incomplete. The complete question is :

Let X be a random variable with probability mass function

P(X =1) =1/2, P(X=2)=1/3, P(X=5)=1/6

(a) Find a function g such that E[g(X)]=1/3 ln(2) + 1/6 ln(5). You answer should give at least the values g(k) for all possible values of k of X, but you can also specify g on a larger set if possible.

(b) Let t be some real number. Find a function g such that E[g(X)] =1/2 e^t + 2/3 e^(2t) + 5/6 e^(5t)

Solution :

Given :

$P(X=1)=\frac{1}{2}, P(X=2)=\frac{1}{3}, P(X=5)=\frac{1}{6}$

a). We know :

$E[g(x)] = \sum g(x)p(x)$

So, $g(1).P(X=1) + g(2).P(X=2)+g(5).P(X=5) = \frac{1}{3} \ln (2) + \frac{1}{6} \ln(5)$

$g(1).\frac{1}{2} + g(2).\frac{1}{3}+g(5).\frac{1}{6} = \frac{1}{3} \ln (2) + \frac{1}{6} \ln (5)$

Therefore comparing both the sides,

$g(2) = \ln (2), g(5) = \ln(5), g(1) = 0 = \ln(1)$

$g(X) = \ln(x)$

Also, $g(1) =\ln(1)=0, g(2)= \ln(2) = 0.6931, g(5) = \ln(5) = 1.6094$

b).

We known that $E[g(x)] = \sum g(x)p(x)$

∴ $g(1).P(X=1) +g(2).P(X=2)+g(5).P(X=5) = \frac{1}{2}e^t+ \frac{2}{3}e^{2t}+ \frac{5}{6}e^{5t}$

$g(1).\frac{1}{2} +g(2).\frac{1}{3}+g(5).\frac{1}{6 }= \frac{1}{2}e^t+ \frac{2}{3}e^{2t}+ \frac{5}{6}e^{5t}$$

Therefore on comparing, we get

$g(1)=e^t, g(2)=2e^{2t}, g(5)=5e^{5t}$

∴ $g(X) = xe^{tx}$

7 0

2 years ago