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Allushta [10]
3 years ago
7

A box can hold 9 cookies. A baker has 878 cookies that she wants to put equally into boxes. How many cookies will be in the last

box that isn't full?
Mathematics
2 answers:
Fofino [41]3 years ago
7 0
6 cookies will be in the last box that would not be full.
divide 679 by 9+97 withe the remainder
97 times 9=873
879-973 is 6
miskamm [114]3 years ago
4 0
97 boxes will be full and the box that isn't full will have 5 cookies in it
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I wasnt in school for the lesson and now i dont know what to do
Wittaler [7]
I hope this helps you



1=3=180-68=112


2=68
7 0
4 years ago
How do i write and solve for this inequality
ExtremeBDS [4]

9514 1404 393

Answer:

  -7/4 < x < 8

Step-by-step explanation:

The value of y can be determined from the sum of the angles, so you know each of the angles exactly. That means you know the ratio of side lengths exactly, which lets you solve for x exactly.

__

Setting that aside, we observe that angle C is greater than angle A, so side AB will be longer than side BC.

  3x +15 > 4x +7

  8 > x . . . . . . . . . . subtract 3x+7

We also know that the lengths of these sides must be positive. Since BC is the shorter side, we require ...

  4x +7 > 0

  4x > -7

  x > -7/4

So, the allowable values of x are ...

  -7/4 < x < 8

_____

<em>More complete solution</em>

If we read the figure correctly, the sum of angles is ...

  (2y +12) +(4y +12) +(y -18) = 180

  7y +6 = 180

  y = (180 -6)/7 = 24 6/7°

Then (in degrees) ...

  ∠A = 2(24 6/7) +12 = 61 5/7, and ∠C = 4(24 6/7) +12 = 111 3/7

The Law of Sines tells us ...

  AB/sin(C) = BC/sin(A)

  sin(A)(3x +15) = sin(C)(4x+7)

  x(4sin(C) -3sin(A)) = 15sin(A) -7sin(C)

  x = (15sin(A) -7sin(C))/(4sin(C) -3sin(A))

  x ≈ 6.1872652

4 0
3 years ago
SOMEBODY, ANY BODY, HELP ME WITH THIS PROBLEM PLS!!!!!
Crazy boy [7]
Because I wanna say sorry if I'm wrong

5 0
3 years ago
Read 2 more answers
What is the price of a $500 tv after a 20% discount?
Hunter-Best [27]
It’s going to be 400 dollars, 20percent is 100 dollars
8 0
3 years ago
A chemical supply company currently has in stock 100 lb of a certain chemical, which it sells to customers in 5-lb batches. Let
rewona [7]

Complete Question:

A chemical supply company currently has in stock 100 lb of a certain chemical, which it sells to customers in 5-lb batches. Let X = the number of batches ordered by a randomly chosen customer, and suppose that X has the pmf shown in the table below. Compute E(X) and V(X).

Then compute the expected number of pounds left after the next customer’s order is shipped and the variance of the number of pounds left.

NOTE: The pmf is shown in the file attached

Answer:

Expected number of batches, E(X) = 2.0 batches

Variance of batches, V(X) = 0.8 batches²

Expected number of pounds left, E(Y) = 90 lb

Variance of pounds left, V(Y) = 20 lb²

Step-by-step explanation:

The company has 100 batches and it sells to customers in 5 lb batches. The number of pounds left after each customer is filled will be modeled by the mathematical formula: Y = 100 - 5X

The expected value of X, E(X) can be calculated using the formula:

E(X) = \sum x p(x)

E(X) = (1*0.3) + (2*0.5) + (3*0.1) + (4*0.1)

E(X) = 0.3 + 1 + 0.3 + 0.4

E(X) = 2.0

Expected number of batches, E(X) = 2.0 batches

The variance of X, V(X) can be calculated using the formula:

V(X) = E(X^2) - [E(X)]^2\\ E(X^2) = \sum x^2 p(x)\\  E(X^2) = (1^2 *0.3) + (2^2 * 0.5) + (3^2 * 0.1) + (4^2 * 0.1)\\  E(X^2) = 0.3 + 2 + 0.9 + 1.6\\ E(X^2) = 4.8\\V(X) = 4.8 - 2^2\\V(X) = 0.8

Variance of batches, V(X) = 0.8 batches²

Y = 100 - 5X

E(Y) = E(100 - 5X)

E(Y) = 100 - 5E(X)

E(Y) = 100 - 5(2)

E(Y) = 90 lb

Expected number of pounds left, E(Y) = 90 lb

Variance of the number of pounds left:

V(Y) = V(100 - 5X)

V(Y) = V(100) + V(-5X)

V(100) = 0

V(Y) = (-5)² V(X)

V(Y) = 25 * 0.8

V(Y) = 20

Variance of pounds left, V(Y) = 20 lb²

3 0
3 years ago
Read 2 more answers
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