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3 years ago

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The math club is selling gift wrap for a funraiser. They sold 45 rolls of soild wrapping paper at $4 each and rolls of patterned

wrapping paper at $5 each. If they made $265, how many rolls of patterned wrapping paper did they sell

1 answer:

Lunna [17]3 years ago

7 0

4 * 45 = 180

265 - 180 = 85

85/5 = 17

They sold 17 rolls of patterned wrapping paper.

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If y varies directly with X and y =10 and X=4 write the direct linear variation equation .

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Y= kx (k being a constant)
10= k × 4 k=5/2

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3 years ago

Help please will give brainliest due at 4:00 pm

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2 years ago

Sophia determined that the expression Negative (4 x minus 5) + 2 (x minus 3) is equivalent to –2x – 5 by using the steps below.

CaHeK987 [17]

Answer:Hope This Helps ☺️

Step-by-step explanation:

She is not correct because she did not substitute the same number in both expressions in Step 1

Step-by-step explanation:

CASE 1: substitute 1 for x to both sides of the equations

L.H.S

-(4x-5)+2(x-3)

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Hence for x= 1

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Because -3 ≠ -7

CASE 2: substitute -1 for x to both sides of the equations

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Answer:

She is not correct because she did not substitute the same number in both expressions in Step 1

3 0

3 years ago

Read 2 more answers

Ninety-one percent of products come off the line within product specifications. Your quality control department selects 15 produ

Allisa [31]

Answer:

Probability of stopping the machine when $X < 9$ is 0.0002

Probability of stopping the machine when $X < 10$ is 0.0013

Probability of stopping the machine when $X < 11$ is 0.0082

Probability of stopping the machine when $X < 12$ is 0.0399

Step-by-step explanation:

There is a random binomial variable $X$ that represents the number of units come off the line within product specifications in a review of $n$ Bernoulli-type trials with probability of success $0.91$ . Therefore, the model is ${15 \choose x} (0.91) ^ {x} (0.09) ^ {(15-x)}$ . So:

$P (X < 9) = 1 - P (X \geq 9) = 1 - [{15 \choose 9} (0.91)^{9}(0.09)^{6}+...+{ 15 \choose 15}(0.91)^{15}(0.09)^{0}] = 0.0002$

$P (X < 10) = 1 - P (X \geq 10) = 1 - [{15 \choose 10}(0.91)^{10}(0.09)^{5}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0013$

$P (X < 11) = 1 - P (X \geq 11) = 1 - [{15 \choose 11}(0.91)^{11}(0.09)^{4}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0082$

$P (X < 12) = 1- P (X \geq 12) = 1 - [{15 \choose 12}(0.91)^{12}(0.09)^{3}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0399$

Probability of stopping the machine when $X < 9$ is 0.0002

Probability of stopping the machine when $X < 10$ is 0.0013

Probability of stopping the machine when $X < 11$ is 0.0082

Probability of stopping the machine when $X < 12$ is 0.0399

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3 years ago

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Talja [164]

(1, 2.50)
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3 years ago