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3 years ago

I NEED HELP ON THIS!!!!!

Mathematics

1 answer:

seropon [69]3 years ago

8 0

#1 pipers pepperoni pieces- $0.41
noras nummy nuggets- $0.34
noras nummy nuggets is better
#2 +65?

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26 x −5 pleae help me i have no idea

Law Incorporation [45]

Answer:

-125

Step-by-step explanation:

negative one twent five

5 0

3 years ago

What is the value of m in the equation? 3(18+ 2) = 4m A.15 B.14 C.6 D.60

Yuki888 [10]

The value of m is D.15

4 0

3 years ago

Read 2 more answers

Place each of the digits 1, 3, 5, and 6 in the blanks provided to create an expression whose value is 1/10

kompoz [17]

Answer:

$\frac{1}{5}\times \frac{3}{6}=\frac{3}{30}$

Step-by-step explanation:

From the picture attached,

Multiplication of two fractions gives the result as $\frac{1}{10}$ .

This shows that the denominator of the fractions when multiplied with each other results a number that will be the multiple of 10.

Therefore, from the given numbers,

5 × 6 = 30

Rest two numbers be placed as the numerator of the fractions.

$\frac{1}{5}\times \frac{3}{6}=\frac{3}{30}$

$=\frac{1}{10}$

Therefore, $\frac{1}{5}\times \frac{3}{6}=\frac{1}{10}$ will be the answer.

3 0

3 years ago

Suppose the number of insect fragments in a chocolate bar follows a Poisson process with the expected number of fragments in a 2

leonid [27]

Answer:

a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b)0.6004

c)19.607

Step-by-step explanation:

Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2

X ~ Poisson(A) where $\lambda = \frac{10.2}{200} = 0.051$

a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar

$\frac{1}{4} \times 200 = 50$

50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55

So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment

Let X denotes the number of grams to be eaten before another fragment is detected.

$P(X>10)= e^{-\lambda \times x}= e^{-0.051 \times 10}= e^{-0.51}=0.6004$

c)The expected number of grams to be eaten before encountering the first fragments :

$E(X)=\frac{1}{\lambda}=\frac{1}{0.051}=19.607 gram$ s

7 0

3 years ago

A town accumulated 2 inches of snow and the depth is increasing by 6 inches every hour. A nearby town ha accumulated 9 inches of

Svetach [21]

Answer: it will take 2.33 hours for the snow fall in both of the towns be equal.

Step-by-step explanation:

Let x represent the number of hours that it will take for the snow fall in both of the towns be equal.

A town accumulated 2 inches of snow and the depth is increasing by 6 inches every hour. This means that in x hours, the depth of snow in the town would be

6x + 2

A nearby town has accumulated 9 inches of snow and the depth is increasing by 3 inches every hour. This means that in x hours, the depth of snow in the town would be

3x + 9

For the snow fall in both of the towns be equal, the number of hours would be

6x + 2 = 3x + 9

6x - 3x = 9 - 2

3x = 7

x = 7/3 = 2.33 hours

8 0

3 years ago