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2 years ago

I need help to these answers asap! 20 points!

Mathematics

2 answers:

sashaice [31]2 years ago

8 0

Answer:

it would be -3 for letter B.

Step-by-step explanation:

why okay reduce the number so it would be 9/4 ÷ 3/4 if that makes sense

iragen [17]2 years ago

5 0

Answer:

Step-by-step explanation:

a is o.53 repeated

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Help plz i give brainliest

Natasha_Volkova [10]

Answer: Hope this helps :)

$\frac{1}{5} b^{27}$

<h2><u><em>Mark me as Brainliest please</em></u></h2>

6 0

2 years ago

Read 2 more answers

PLEASE HELP timed exam pls

3241004551 [841]

Answer:

helping hand help to write good

5 0

2 years ago

23 and 24 I need help please

Gnoma [55]

Answer:

They should purchase the snack with the purple box. This is the answer for 23, i don't know what 24 is sorry.

Step-by-step explanation:

24÷3=8

8×5=40

so the first snacks would cost them $40.

then:

24÷4=6

6×6=36

so the second snacks would cost them $36

5 0

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.

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

5 0

3 years ago

PLEASE HELP ME! Images below & I give Brainliest

neonofarm [45]

Answer:

Graph C with the green dots.

Step-by-step explanation:

y = 0.5x + 5 is a linear equation, so we can rule out A because it is not straight.

5 represents the y-intercept, so we rule out B because that graph doesn't intercept the y-axis at (0,5).

Graph C represents a straight line that passes through (0, 5) & has a 0.5 slope, so that is our answer.

Hope this helps. :0)

4 0

2 years ago