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3 years ago

1.)Each sandwich costs $.02 per cubic inch to make. What is the cost of each sandwich?

1 answer:

7 0

Part 1) Each sandwich costs $.02 per cubic inch to make. What is the cost of each sandwich?

volume for cube sandwich is 125 in³
volume for pyramid sandwich is 41.67 in³

cost of each cube sandwich=0.02*125-----> $2.5

cost of each pyramid sandwich=0.02*41.67-----> $0.83

the answers Part 1) are
a) cost of each cube sandwich is $2.5
b) cost of each pyramid sandwich is $0.83

Part 2) If each sandwich sells for $4.00 and Mr. Sammie sells 100 Cube sandwiches and 100 Pyramid Sandwiches, how much profit would Mr. Sammie make?

we know that
Profit=Sells-Costs

step 1
find the sells
Sells=200*$4-----> $800

step 2
find the costs
Costs=100*$2.5+100*$0.83----> $333

step 3
find the profit
Profit=Sells-Costs
Profit=$800-$333-----> $467

the answer Part 2) is
the profit is $467

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In 1999 bus fare in atlanta was 4.70.In 1979 the fare was 2/5 of the fare in 1999 what was the fare in 1979

Gennadij [26K]

Answer : 1.96 was the fare in 1979.

Explanation :

Bus fare in Atlanta in 1999 =4.70

$\text{As we have given that in 1979 the fare was } \frac{2}{5} \text{ of the fare in 1999.}$

$\text{Bus fare in Atlanta in 1979=} \frac{2}{5}\times 4.70$

Bus fare in Atlanta in 1979 is given by

$\frac{2\times 4.70}{5}\\\\=\frac{9.80}{5}\\\\=1.96$

∴ 1.96 was the fare in 1979.

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3 years ago

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The lifetime of a certain type of battery is normally distributed with mean value 15 hours and standard deviation 1 hour. There

KIM [24]

Answer:

If the lifetime of batteries in the packet is 40.83 hours or more then, it exceeds for 5% of all packages.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 15

Standard Deviation, σ = 1

Sample size = 4

Total lifetime of 4 batteries = 40 hours

We are given that the distribution of lifetime is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling:

$\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{1}{\sqrt4} = 0.5$

We have to find the value of x such that the probability is 0.05

P(X > x) = 0.05

$P( X > x) = P( z > \displaystyle\frac{x - 40}{0.5})=0.05$

$= 1 -P( z \leq \displaystyle\frac{x - 40}{0.5})=0.05$

$=P( z \leq \displaystyle\frac{x - 40}{0.5})=0.95$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 40}{0.5} = 1.64\\x = 40.825 \approx 40.83$

Hence, if the lifetime of batteries in the packet is 40.83 hours or more then, it exceeds for 5% of all packages.

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3 years ago

Ethan sold 2 3 gallon of lemonade. Kayla sold some lemonade too. Together, they sold 1 1 4 gallons. Who sold more lemonade, Etha

dezoksy [38]

Answer:

Kayla sold more lemonade.

Step-by-step explanation

All you do to solve this problem is subtract 23 from 114.

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3 years ago

Among cases of heart pacemaker malfunctions, were found to be caused by firmware, which is software programmed into the device.

alex41 [277]

Complete question is;

Among 8834 cases of heart pacemaker malfunctions, 504 were found to be caused by firmware, which is software programmed into the device. If the firmware is tested in three different pacemakers randomly selected from this batch of 8834 and the entire batch is accepted if there are no failures, what is the probability that the firmware in the entire batch will be accepted? Is this procedure likely to result in the entire batch being accepted?

Answer:

P(All three are not caused by firmware) = 83.84%

Probability that the entire batch will be accepted = 0.8384

Step-by-step explanation:

We are told that out of the 8834 cases of heart pacemaker malfunctions, 504 were found to be caused by firmware.

Thus,

Cases not caused by firmware = 8834 - 504 = 8330

So, probability of the first case not being affected by firmware is;

P(first case not caused by firmware) = 8330/8834

Also,

Probability of second case not being affected by firmware is given as;

P(second case not caused by firmware|first case not affected by firmware) = 8329/8833

Similarly,

Probability of third case not being affected by firmware is given as;

P(third case not caused by firmware|first and second not caused by firmware) = 8328/8832

Now, looking at the 3 Probabilities gotten, it is obvious that the events are not independent because the probability of occurence of one event depends on the probability of occurence of the other event.

Thus, we will make use of the general multiplication rule which is;

P(A & B) = P(B) × P(A|B)

Thus;

P(All three not caused by firmware) = P(first case not caused by firmware) × P(second case not caused by firmware|first case not affected by firmware) × P(third case not caused by firmware|first and second not caused by firmware)

Plugging in the relevant values, we have;

P(All three not caused by firmware) = (8330/8834) × (8329/8833) × (8328/8832)

P(All three are not caused by firmware) = 0.83840506679 ≈ 83.84%

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3 years ago

Complete the table of equivalent ratios. 3:2 ?:4 9:6 12:8 15:10

Vera_Pavlovna [14]

Answer:

6:4

Step-by-step explanation:

well on one side your adding 3 to the next ratio

so on the other side your doing the same of adding 2 to the next ratio

also they all reduce down to 3:2

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3 years ago

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