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sineoko [7]
3 years ago
8

. If F, G, and H are the midpoints of the sides of ∆CDE, FG = 9, GH = 7, and CD = 24, find each measure.

Mathematics
1 answer:
Lady_Fox [76]3 years ago
7 0

WHERE'S THE LEAK MA'AM

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Solve for X. Show your work<br><br> 2x + 6 = x - 5
irina1246 [14]

2x + 6 = x - 5

2x - x = -5 - 6

x = -11 ← the end

8 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
A survey of 280 homeless persons showed that 63 were veterans. Construct a 90% confidence interval for the proportion of homeles
MA_775_DIABLO [31]

Answer:

[0.184, 0.266]

Step-by-step explanation:

Given:

Number of survey n =280

Number of veterans = 63

Confidence interval = 90%

Computation:

Probability of veterans = 63/280

Probability of veterans =0.225

a=0.1

Z(0.05) = 1.645 (from distribution table)

Confidence interval = 90%

So,

p ± Z*√[p(1-p)/n]

0.225 ± 1.645√(0.225(1-0.225)/280)

[0.184, 0.266]

4 0
3 years ago
Vernon tossed a coin 20 times. The results were 8 head s and 12 tails. What is the experimental probability of tossing heads?
solmaris [256]

Answer:

2/5

Step-by-step explanation:

The experimental probability is

P (heads) = number of heads/ total tosses

                = 8/20

                 = 2/5

8 0
3 years ago
Which is the side length of a cube with a volume of 2744 cm3?<br><br> Please help
Elis [28]
\sf~V=a^3

Plug in what we know:

\sf2744=a^3

Find the cube root of both sides:

\sf~a=\sqrt[3]{2744}

Simplify:

\sf~a=\boxed{\sf14}
3 0
2 years ago
Read 2 more answers
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