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3 years ago

12

at the end of a Pizza party 1/3 of a cheese pizza and 1/6 of a pepperoni pizza or left brand ate 1/2 of all the leftover pizza w

hat fractional part of a whole pizza did Brand eat

1 answer:

Readme [11.4K]3 years ago

5 0

Answer:

Half?

I’m not sure myself

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The gross weight of a truck containing 30 pigs is 8 tons 1,123 pounds and 6 ounces. The average weight of one pig is 95 pounds 1

irakobra [83]

Answer: The net weight of the pigs of the truck is 1 ton 868 pounds 12 ounces.

The weight of the empty truck is 7 tons 254 pounds 10 ounces.

Step-by-step explanation:

Since we have given that

The gross weight of a truck containing 30 pigs = 8 tons 1123 pounds 6 ounces

Average weight of one pig = 95 pounds 10 ounces

Now,

We know that

$1\text{ pound}=16\text{ ounces}$

So,

$\text{Gross weight of truck containing 30 pigs}\\=8\text{ tones }1123\times 16+6\text{ ounces}\\=8\text{ tons }17974\text{ ounces}$

Similarly,

$\text{ average weight of one pig }\\=95\times 16+10\text{ ounces}\\=1530\text{ ounces}$

So, Net weight of the pigs in the truck

$1530\times 30\\ =45900\text{ ounces}\\\\=\frac{45900}{16}\\\\=2868\text{ pounds }12\text{ ounces}\\\\=1\text{ ton }868\text{ pounds }12\text{ ounces }$

So, the net weight of the pigs of the truck is 1 ton 868 pounds 12 ounces.

Now, we'll find the weight of the empty truck is given by

$\text{Gross weight of a truck containing} -\text{ Net weight of pigs in the truck}\\=8\text{ tons}1123\text{ pounds}6\text{ ounces}-1\text{ton }868\text{ pounds }12\text{ounces}\\=(8\times 2000\times 16+1123\times 16+6)-(1\times 2000\times 16+868\times 16-12)\\=273974\text{ ounces}-45900\text{ ounces}\\=228074\text{ounces }\\=7\text{ tons }254\text{ pounds }10\text{ ounces}$

Hence, the weight of the empty truck is 7 tons 254 pounds 10 ounces .

4 0

3 years ago

Please help me! (this is from khan academy 6th grade math!)

Svetach [21]

Answer:

1st → 1st

2nd → 3rd

3rd → 2nd

Step-by-step explanation:

just divide the first set of student wishes to the second set of students in the same question. the 1st for example

x-ray to all students is 6/ 36= 1/6 = 1:6

for all students, you add all the students in the table

8 0

1 year ago

Two different suppliers, A and B, provide a manufacturer with the same part. All supplies of this part are kept in a large bin.

koban [17]

Answer:

The probability of selecting a non-defective part provided by supplier A is 0.807.

Step-by-step explanation:

Let <em>A</em> = a part is supplied by supplier A, <em>B</em> = a part is supplied by supplier B and <em>D</em> = a part is defective.

<u>Given</u>:

P (D|A) = 0.05, P(D|B) = 0.09

A supplies four times as many parts as B, i.e. n (A) = 4 and n (B) = 1.

Then the probability of event <em>A</em> and <em>B</em> is:

$P(A)=\frac{n(A)}{n(A)+N(B)}= \frac{4}{4+1}=0.80\\P(B)\frac{n(B)}{n(A)+N(B)}= \frac{1}{4+1}=0.20$

Compute the probability of selecting a defective product:

$P(D)=P(D|A)P(A)+P(D|B)P(B)\\=(0.05\times0.80)+(0.09\times0.20)\\=0.058$

The probability of selecting a non-defective part provided by supplier A is:

$P(A|D')=\frac{P(D'|A)P(A)}{P(D')} = \frac{(1-P(D|A))P(A)}{1-P(D)}\\=\frac{(1-0.05)\times0.80}{(1-0.058)}\\ =0.80679\\\approx0.807$

Thus, the probability of selecting a non-defective part provided by supplier A is 0.807.

5 0

3 years ago

Jayden's new puppy weighed 11 1/4 pounds at 4 weeks old. At 16 weeks old, the puppy weighed 1 2/3 times more. How much did Jayde

Whitepunk [10]

<h2><u>Answer:</u></h2><h2><u>18 3/4 pounds</u></h2>

<u></u>

7 0

3 years ago

Last year Frank had a total income of $58,800. He sold a house and made a profit of $27,940. He also had monthly income of $80 f

xeze [42]

I think the answer would have to be D. $0.90.

59,970-58,800=1,170
1,170/52=22.5/25=0.90

3 0

3 years ago

Read 2 more answers