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3 years ago

12

There are 6 2/3 pounds of walnuts in a container, which will be divided equally into containers that hold 1 1/2 pounds. This wou

ld fill ____ and 4/9 containers.
Write a number, not a decimal or fraction.

2 answers:

Rudiy273 years ago

8 0

Answer: It would fill 4 and 4/9

Sloan [31]3 years ago

4 0

Answer:

This would fill 4 and 4/9 containers.

hope it helps!!

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The price paid for a $15 shirt after a 45% discount is applied

Tamiku [17]

Answer:

ur mom

Step-by-step explanation:

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3 years ago

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What is the common ratio of 40,12,3.6

slava [35]

Answer:

0.3

Step-by-step explanation:

Divide the second term (12) with the first term (40), and you'll get 0.3

This is correct because you get the terms in the correct order when multiplying:
(First term: 40. Common ratio: 0.3)
40 x 0.3 = <u>12</u>
<u>12</u> x 0.3<em> </em>= <em>3.6</em>

40, <u>12</u>, <em>3.6</em>

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2 years ago

Given: circle k(O), m∠P=95°, m∠J=110°, m∠LK=125°<br> Find: m∠PJ

IRINA_888 [86]

Answer:

The measure of the arc PJ is $75\°$

Step-by-step explanation:

step 1

Find the measure of angle L

we know that

In a inscribed quadrilateral opposite angles are supplementary

so

$m$

we have

$m$

substitute

$m$

$m$

step 2

Find the measure of arc KJ

we know that

The inscribed angle measures half that of the arc comprising

so

$m$

substitute the values

$95\°=\frac{1}{2}(125\°+arc\ KJ)$

$190\°=(125\°+arc\ KJ)$

$arc\ KJ=190\°-125\°=65\°$

step 3

Find the measure of arc PJ

we know that

The inscribed angle measures half that of the arc comprising

so

$m$

substitute the values

$70\°=\frac{1}{2}(65\°+arc\ PJ)$

$140\°=(65\°+arc\ PJ)$

$arc\ PJ=140\°-65\°=75\°$

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3 years ago

Which system of equations does not have a real solution

Aloiza [94]

Answer:

the correct answer is y = x 2 + 3x - 5 and x + y = -10

Step-by-step explanation:

5 0

3 years ago

A donut store has 11 different types of donuts. You can only buy a bag of 3 of them, where each donut has to be of a different t

MakcuM [25]

Answer:

$165$ .

Step-by-step explanation:

Since repetition isn't allowed, there would be $11$ choices for the first donut, $(11 - 1) = 10$ choices for the second donut, and $(11 - 2) = 9$ choices for the third donut. If the order in which donuts are placed in the bag matters, there would be $11 \times 10 \times 9$ unique ways to choose a bag of these donuts.

In practice, donuts in the bag are mixed, and the ordering of donuts doesn't matter. The same way of counting would then count every possible mix of three donuts type $3 \times 2 \times 1 = 6$ times.

For example, if a bag includes donut of type $x$ , $y$ , and $z$ , the count $11 \times 10 \times 9$ would include the following $3 \times 2 \times 1$ arrangements:

$xyz$ .
$xzy$ .
$yxz$ .
$yzx$ .
$zxy$ .
$zyx$ .

Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count $11 \times 10 \times 9$ by $3 \times 2 \times 1 = 6$ to find the actual number of donut combinations:

$\begin{aligned} \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of $3$ objects from a set of $11$ distinct objects:

$\begin{aligned}\begin{pmatrix}11 \\ 3\end{pmatrix} &= \frac{11 !}{(11 - 3)! \times 3 !} \\ &= \frac{11 !}{8 ! \times 3 !} \\ &= \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

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2 years ago