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3 years ago

You have 9 1/2 on a poster for a row of photos. Each photo is 2 3/4 inches wide. How many photos can you fit?

Mathematics

1 answer:

Sever21 [200]3 years ago

4 0

Answer:

i wanna say 3

Step-by-step explanation:

i believe its 3 because if you add that 1/2 to the 3/4 you could have a 3 and 3 divided by 9 is 3 (forgive me if incorrect)

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Hi can anyone help me with this please

kupik [55]

Answer:

sure

Step-by-step explanation:

she multiplied it 3 times, so, i dont think 8 should be negative.

6 0

3 years ago

Read 2 more answers

What do we do to both sides of the equation? X-8=3

Alik [6]

Answer:

$\huge\fbox\red{a}\huge\fbox\orange{n} \huge\fbox\pink{s}\huge\fbox\green{w} \huge\fbox\blue{e}\huge\fbox\purple{r}\$

We'll add 8 to the both side of the equation on order to get the Value of x
x -8+8 = 3+8
x = 11

~ʆᵒŕ∂ཇꜱꜹⱽẻⱮë

5 0

2 years ago

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Find the value of x in the equation below.<br> 17 = 14 +2

ozzi

Answer:

Step-by-step explanation:

8 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

3 years ago

15+x<24 what is the solution to this inequality

ra1l [238]

Move all terms not containing x to the right side of the inequality. x < 9

5 0

3 years ago