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Yakvenalex [24]
3 years ago
9

Janie ordered boxed lunches for a student advisory committee meeting. Each lunch costs $4. The total cost of the lunches is $47,

including a $7 delivery fee. Write and solve an equation to find x, the number of boxed lunches Janie ordered.
Mathematics
2 answers:
ladessa [460]3 years ago
7 0

Hello, 928747!

So we are trying to find how many boxed lunches Janie ordered.

Each box is $4.00 and the total cost is $47.00 + delivery fee of $7.00

So to find this, we will first subtract the delivery fee because that does not count towards the lunches.

47 - 7 = 40

So the boxed lunches cost $40.00

To find how many Janie ordered, we will divide 4 by 40

40 ÷ 4 = 10

So Janie bought 10 boxed lunches.

The equation would be (47 - 7) ÷ 4

#LearnWithbrainly

<u>brainly.com/question/21306300</u>

Elodia [21]3 years ago
6 0

Answer:

10 lunches

Step-by-step explanation:

$47-$7=$40

40/4=10

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Suppose a random variable x is best described by a uniform probability distribution with range 22 to 55. Find the value of a tha
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(b) The value of <em>a</em> is 38.17.

(c) The value of <em>a</em> is 26.95.

(d) The value of <em>a</em> is 25.63.

(e) The value of <em>a</em> is 12.06.

Step-by-step explanation:

The probability density function of <em>X</em> is:

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(a)

Compute the value of <em>a</em> as follows:

P(X\leq a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.95\times 33=[x]^{a}_{22}\\\\31.35=a-22\\\\a=31.35+22\\\\a=53.35

Thus, the value of <em>a</em> is 53.35.

(b)

Compute the value of <em>a</em> as follows:

P(X< a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.49\times 33=[x]^{a}_{22}\\\\16.17=a-22\\\\a=16.17+22\\\\a=38.17

Thus, the value of <em>a</em> is 38.17.

(c)

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P(X\geq  a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.85=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.85\times 33=[x]^{55}_{a}\\\\28.05=55-a\\\\a=55-28.05\\\\a=26.95

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(d)

Compute the value of <em>a</em> as follows:

P(X\geq  a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.89=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.89\times 33=[x]^{55}_{a}\\\\29.37=55-a\\\\a=55-29.37\\\\a=25.63

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P(1.83\leq X\leq  a)=\int\limits^{a}_{1.83} {\frac{1}{33}} \, dx \\\\0.31=\frac{1}{33}\cdot \int\limits^{a}_{1.83} {1} \, dx \\\\0.31\times 33=[x]^{a}_{1.83}\\\\10.23=a-1.83\\\\a=10.23+1.83\\\\a=12.06

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