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2 years ago

I nene help with this please. It a quiz

Mathematics

1 answer:

Vladimir [108]2 years ago

8 0

58 if you add the two numbers then subtract the answer from 180 that’s the answer

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In general, the intercept of the function F(X) = a•b^x is the point

Murljashka [212]

Answer:

$(0,a)$

Step-by-step explanation:

Given

$f(x) = ab^x$

Required

Determine the intercept

The intercept is at point: $x = 0$

So, we have:

$f(0) = ab^0$

$f(0) = a*1$

$f(0) = a$

So, the intercept is at point $(0,a)$

<em>None of the options is true</em>

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2 years ago

Teachers drive a mini-van. what percentage of the teachers drive a mini-van? (6.5B*) There are 60 teachers at Smith Middle Schoo

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20 o/o fndjdndbbbbbndndn

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3 years ago

4|m−n| if m=−7 and n=2 i really need to know what this is asap please

vekshin1

Answer:

Step-by-step explanation:

Plug in -7 as m and 2 as n into the expression:

4 | m - n |

4 | -7 -2 |

Solve:

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2 years ago

Select all the expressions that have a value of 40 when x = 10.8.

natta225 [31]

Answer:

Step-by-step explanation:

B. =16x+288x

C. = 1 18 x2+33.52

1752

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162

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5 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