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2 years ago

I have one more question this is the last one im asking and again don't just put any old answer or youl be reported.

Mathematics

2 answers:

ladessa [460]2 years ago

5 0

Answer:

multiply

Step-by-step explanation:

pls don't report if I a wrong!

vladimir1956 [14]2 years ago

3 0

Your work is blurry bro sorry

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Your math test scores are 100, 100, 90, 93, and 52. what is the mean of your scores?

mixer [17]

435.
Add em all up and you’ll get 435.

7 0

2 years ago

Find the product. y4 · y5

Alexandra [31]

I would assume this to be equal to y⁴ ₓ y⁵

= y⁴ ⁺ ⁵, By law of indices.

= y⁹

I hope this helped.

8 0

3 years ago

PLEASE ANSWER!!! You are helping wish some repairs at home. you drop a hammer and it hits the floor at a speed of 12 feet per se

Andru [333]

Answer:

h = 2.25 m

Step-by-step explanation:

Given:

Speed = v = 12 ft/sec

Acceleration due to gravity = g = 32 ft/sec²

Required:

Height = h = ?

Formula:

v = $\sqrt{2gh}$

Solution:

For h, the formula will be:

=> h = $\frac{v^2}{2g}$

=> h = $\frac{(12)^2}{2(32)}$

=> h = $\frac{144}{64}$

=> h = 2.25 m

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

1) how many significant digits are there in the sum of 45.10 and 102.203?

Sergio039 [100]

Oi :)

1) There are 3 significant digits
$45.10+102.203= 147.303$

2)
15 Mi/h = 22 ft/s

3)
$C= 72*m* \frac{3}{4} \\ \\ C= 72*81* \frac{3}{4} \\ \\ C= \frac{17496}{4} \\ \\ C=4374$

4)
There are 4 significant digitis. Leter C

7 0

2 years ago