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3 years ago

Two sailboats are traveling along the same path away

Mathematics

1 answer:

kodGreya [7K]3 years ago

4 0

Likely B

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She earns $3.10 for each hour she works and $2.30 for each bushel of apples she picks. Her goal is to earn at least $100 this we

nika2105 [10]

The inequality would be 3.1h + 2.3b ≥100

Step-by-step explanation:

Earning of Tammy for each hour of work= $3.1

Total hours she worked= h

Earning for picking each bushel of apple= $2.3

Total bushel she picks= b

Her goal = minimum of $100

Hence total earning from working= earning from each hour* no of hours worked

Total earning= 3.1h

Similarly, total earning from picking up apple bushels= earning from picking one bushel*no of bushels picked

Total earning= 2.3b

According to question total earning must be greater than $100

Hence, 3.1h + 2.3b ≥100

3 0

3 years ago

What is the equation of the line that passes through the points 15, nine and -2, nine

olasank [31]

Answer:

y=9

Step-by-step explanation:

y=mx+c

gradient of a horizontal line is 0

y=0x+9

y=9

5 0

3 years ago

What is the explicit formula for this sequence? -5,-3,-1,1,3

Y_Kistochka [10]

Answer:

Step-by-step explanation:

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5 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 cube has edge length 3 in. What is its volume in inches cubed?

Nadusha1986 [10]

Answer:

$Volume = 33 = 27 inches^{3}$

Can u gimme brain plz!

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5 0

2 years ago