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3 years ago

WHAT IS THE ANSWER TO 0.16PSQUARED-0.0025

Mathematics

1 answer:

Misha Larkins [42]3 years ago

8 0

$(0.16)^2-0.0025=0.0256-0.0025=0.0231$

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grigory [225]

Answer:

m = $\frac{7}{5}$

Step-by-step explanation:

m - $\frac{m-1}{2}$ = 1 - $\frac{m-2}{3}$

Multiply through by 6 ( the LCM of 2 and 3 ) to clear the fractions

6m - 3(m - 1) = 6 - 2(m - 2) ← distribute parenthesis on both sides

6m - 3m + 3 = 6 - 2m + 4 , that is

3m + 3 = - 2m + 10 ( add 2m to both sides )

5m + 3 = 10 ( subtract 3 from both sides )

5m = 7 ( divide both sides by 5 )

m = $\frac{7}{5}$

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3 years ago

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2 years ago

Write √3 x √6 in the form b√2 where b is an integer

bagirrra123 [75]

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3 years ago

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.

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

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3 years ago

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Ghella [55]

The answer is A. life insurance that pays a benefit on the death of the insured and also accumulates a cash value.

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3 years ago

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