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3 years ago

Aiden picked 3/8 of the carrots in his garden. There were 32 carrots in his garden.

2 answers:

34kurt3 years ago

4 0

Hello! I believe the answer is that Aiden picked 12 carrots!
Here's how I got this answer :
3/8 = x/32 , x is the unknown amount of carrots he picked!
Let's take a look at the denominators :
To get to 32 from 8, we multiply by 4.
We can apply this to the numerator! So, :
4(3) = 12, so then the x = 12!

Ann [662]3 years ago

3 0

The answer is 12 carrots:)

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7). John had 9 pieces of gum at the

Dmitry_Shevchenko [17]

Answer:

9-1=8

Step-by-step explanation:

he started with 9 but at the end of the day he had 8. meaning he ate 1 peice

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1 year 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

Please help me...........................

malfutka [58]

Answer:

Step-by-step explanation: because this negative

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2 years ago

Given; D is the Midpoint of CE / Prove DE =1/2CE

aleksley [76]

Answer:

Check explanation

Step-by-step explanation:

Here, we want to make a prove;

Mathematically , since D is the midpoint of CE

Then;

CE = CD + DE

Also, since D splits the line segment into two equal parts as the midpoint, then CD must be equal to DE

I.e CD = DE

Hence, we can express CE as follows;

CE = DE + DE

CE = 2 DE

Divide both sides by 2

CE/2 = DE

Hence; DE = 1/2 CE

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2 years ago

If the range of f (x) = StartRoot m x EndRoot and the range of g (x) = m StartRoot x EndRoot are the same, which statement is tr

vagabundo [1.1K]

Answer:

M IS M

Step-by-step explanation:

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2 years ago