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Julli [10]
3 years ago
13

A restaurant purchased 18 cases of hamburgers with 26 hamburgers in each case. Since that time they sold 73 hamburgers. How many

hamburgers do they have now?
Mathematics
1 answer:
Lady bird [3.3K]3 years ago
7 0

Answer:

395 Burgers.

Step-by-step explanation:

each crate has 26 burgers, if there are 18 crates its 18x26 which gives you 468 burgers. If they sold 73 after that, you subtract. 468-73= 395. They will have 395 burgers.

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Please answer this by solving it
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Answer:

5 5/6

Step-by-step explanation:

Change them into improper fractions.  Multiply the denominator by the whole number and then add the numerator.  Put that number over the denominator.

2 X 3 = 6           3 X 2 = 6

6 + 1 = 7            6 + 1 = 7

7/3                      7/2

New expression:    7/3 + 7/2

Find a common denominator.  Both denominators can go into 6.

14/6 + 21/6 = 35/6

Change this back into a mixed number by dividing.

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You cut a 54-inch rope into four pieces. Twoof the pieces are the same length, a thirdpiece is twice as long as each of the twoe
antiseptic1488 [7]

Let's use the variable x to represent the length of the first and second pieces.

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Adding all four pieces and equating to 54 inches, we have:

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5 0
1 year ago
broderick had 420 dollars in his checking account he made 6 deposits of 35.50 each.He needs to write 4 checks for 310.75 each.Wh
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3 years ago
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Suppose a certain capsule is manufactured so that the dosage of the active ingredient follows the distribution Y ~ N(μ = 10 mg,
lianna [129]

Answer:

(a) Probability that Y falls into the dangerous region is 0.0013.

(b) Probability that the mean Y-bar falls into the dangerous region is 0.00001.

Step-by-step explanation:

We are given that a certain capsule is manufactured so that the dosage of the active ingredient follows the distribution Y ~ N(μ = 10 mg, σ = 1 mg).

A dosage of 13 mg is considered dangerous.

Let Y = <u><em>dosage of the active ingredient </em></u>

The z-score probability distribution for normal distribution is given by;

                                 Z  =  \frac{ Y-\mu}{\sigma} } }  ~ N(0,1)

where, \mu = population mean = 10 mg

            \sigma = standard deviation = 1 mg

(a) Probability that Y falls into the dangerous region is given by = P(Y \geq 13 mg)

       P(Y \geq 13 mg) = P( \frac{ Y-\mu}{\sigma} } } \geq \frac{ 13-10}{1} } } ) = P(Z \geq 3) = 1 - P(Z < 3)  

                                                          = 1 - 0.9987 = <u>0.0013</u>

The above probability is calculated by looking at the value of x = 3 in the z table which has an area of 0.9987.

(b) We are given that a dosage of 13 mg is considered dangerous. And we sample 49 capsules at random.

Let \bar Y = sample mean dosage

The z-score probability distribution for sample mean is given by;

                                 Z  =  \frac{\bar Y-\mu}{\frac{\sigma}{\sqrt{n} } } } }  ~ N(0,1)

where, \mu = population mean = 10 mg

            \sigma = standard deviation = 1 mg

            n = sample of capsules = 49

So, Probability that the mean Y-bar falls into the dangerous region is given by = P(\bar Y \geq 13 mg)

          P(Y \geq 13 mg) = P( \frac{\bar Y-\mu}{\frac{\sigma}{\sqrt{n} } } } } \geq \frac{13-10}{\frac{1}{\sqrt{49} } } } } ) = P(Z \geq 21) = 1 - P(Z < 21)  

                                                             = <u>0.00001</u>

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