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4 years ago

15

a group of mountain climbers begin an expedition with 270 pounds of food. They plan to eat a total of 15 pounds of food per day.

Write and solve a linear equation to find number of days their food will last.

2 answers:

umka2103 [35]4 years ago

5 0

The answer will be 18.

Hope this helped, and please mark brainliest! :)

Crazy boy [7]4 years ago

3 0

The equation would be 270/15 (divide) and the answer would be 18

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Mr. Santiago can buy light fixtures in packages of 12 and light bulbs in packages of 9. He bought the fewest number of light fix

liq [111]

Answer:

3 and 4 packages

Step-by-step explanation:

We need to find the LCM of 12 and 9.

9 = 3*3
12 = 2*2*3

<u>Their LCM is:</u>

LCM(12, 9) = 2*2*3*3 = 36

Number of light fixtures and light bulbs is 36.

<u>Number of packages will be:</u>

36/12 = 3 and
36/9 = 4 respectively.

6 0

3 years ago

Read 2 more answers

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

5 0

4 years ago

Can you guys help with me

Anika [276]

Answer:

x=8

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Of the following points, name all that lie on the same horizontal line?

ArbitrLikvidat [17]

I just graphed all 4 and didn’t see where any are on the same horizontal line. I might be missing something.

7 0

3 years ago

A right triangle has a hypotenuse with a length of 130 cm. One leg has a length of 100 cm. What is the length of the other leg o

saveliy_v [14]

Answer:

B. √6900

Step-by-step explanation:

The square length of hypotenuse is equal to sum of square length of two legs let x represent the other leg

130^2 = 100^2 + x^2

16900 = 10000 + x^2 subtract 10000 from both sides

6900 = x^2 find the root for both sides

x = √6900

3 0

3 years ago