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frutty [35]
3 years ago
13

During the day, the manager of the meat department sold 15.4 pounds of hamburger to one customer and 13.22 pounds to another cus

tomer. When the store opened that morning, he had 35.75 pounds of hamburger to sell. How much hamburger does the manager have left to sell? lb
Mathematics
1 answer:
julsineya [31]3 years ago
7 0
The total amount of hamburger that the manager of the store in the morning = 35.75 pounds
Quantity of hamburger sold to one customer during the day = 15.4 pounds
Quantity of hamburger sold to another customer during the day = 13.22 pounds
Total quantity of hamburger sold
by the manager during the day = (15.4 + 13.22) pounds
                                                  = 28.62 pounds
Then
The quantity of hamburger left in the store to sell = (35.75 - 28.62) pounds
                                                                               = 7.13 pounds
So 7.13 pounds of hamburger is still left to be sold.
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Expressions equal to 10^5
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10 x 10 x 10 x 10 x 10, 20 x 20 x 10, 30 x 20

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Anton [14]

Answer:

The distance between the pirate and the treasure can be found from the following relationship between ΔEAF and ΔABC;

\overline{AE}/\overline{AB} =  \overline{FA}/\overline{AC} = \overline{EF}/\overline{BC}

Step-by-step explanation:

From the question diagram, we have two triangles, ΔEAF and ΔABC;

The ratio of the lengths of the sides \overline{AE}/\overline{AB} =  \overline{FA}/\overline{AC}

∠EAF and ∠BAC are vertical angles, therefore ∠EAF = ∠BAC

Therefore, ΔEAF and ΔABC are similar triangles by Side-Angle-Side, SAS, rule of similarity which states that two triangles that have ratios of a pair of their corresponding sides and the two sides also form equal angles within each triangle, then the two triangles are similar

Therefore, the ratio of the each pair of corresponding sides of the two triangles are equal

We have;

\overline{AE}/\overline{AB} =  \overline{FA}/\overline{AC} = 50 ft. /(100 ft.) =  \overline{EF}/\overline{BC} = \overline{EF}/120 ft.

\overline{EF} = 120 ft. × 50 ft./(100 ft.) = 60 ft.

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4 0
3 years ago
The computers of nine engineers at a certain company are to be replaced. Four of the engineers have selected laptops and the oth
Gala2k [10]

Answer:

(a) There are 70 different ways set up 4 computers out of 8.

(b) The probability that exactly three of the selected computers are desktops is 0.305.

(c) The probability that at least three of the selected computers are desktops is 0.401.

Step-by-step explanation:

Of the 9 new computers 4 are laptops and 5 are desktop.

Let X = a laptop is selected and Y = a desktop is selected.

The probability of selecting a laptop is = P(Laptop) = p_{X} = \frac{4}{9}

The probability of selecting a desktop is = P(Desktop) = p_{Y} = \frac{5}{9}

Then both X and Y follows Binomial distribution.

X\sim Bin(9, \frac{4}{9})\\ Y\sim Bin(9, \frac{5}{9})

The probability function of a binomial distribution is:

P(U=k)={n\choose k}\times(p)^{k}\times (1-p)^{n-k}

(a)

Combination is used to determine the number of ways to select <em>k</em> objects from <em>n</em> distinct objects without replacement.

It is denotes as: {n\choose k}=\frac{n!}{k!(n-k)!}

In this case 4 computers are to selected of 8 to be set up. Since there cannot be replacement, i.e. we cannot set up one computer twice or thrice, use combinations to determine the number of ways to set up 4 computers of 8.

The number of ways to set up 4 computers of 8 is:

{8\choose 4}=\frac{8!}{4!(8-4)!}\\=\frac{8!}{4!\times 4!} \\=70

Thus, there are 70 different ways set up 4 computers out of 8.

(b)

It is provided that 4 computers are randomly selected.

Compute the probability that exactly 3 of the 4 computers selected are desktops as follows:

P(Y=3)={4\choose 3}\times(\frac{5}{9})^{3}\times (1-\frac{5}{9})^{4-3}\\=4\times\frac{125}{729}\times\frac{4}{9}\\  =0.304832\\\approx0.305

Thus, the probability that exactly three of the selected computers are desktops is 0.305.

(c)

Compute the probability that of the 4 computers selected at least 3 are desktops as follows:

P(Y\geq 3)=1-P(Y

Thus, the probability that at least three of the selected computers are desktops is 0.401.

6 0
3 years ago
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