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sashaice [31]
3 years ago
12

15,614+?=68,025 What is the total

Mathematics
1 answer:
mezya [45]3 years ago
8 0

Answer:

52411 that's the answer

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A real estate agent has 19 properties that she shows. She feels that there is a 30% chance of selling any one property during a
netineya [11]

Answer:

P(X \geq 5)=1-P(X

We can find the individual probabilities:

P(X=0)=(19C0)(0.3)^0 (1-0.3)^{19-0}=0.00114

P(X=1)=(19C1)(0.3)^1 (1-0.3)^{19-1}=0.0092

P(X=2)=(19C2)(0.3)^2 (1-0.3)^{19-2}=0.0358

P(X=3)=(19C3)(0.3)^3 (1-0.3)^{19-3}=0.0869

P(X=4)=(19C4)(0.3)^4 (1-0.3)^{19-4}=0.1491

And replacing we got:

P(X \geq 5) = 1-[0.00114+0.009282+0.0358+0.0869+0.149]= 0.7178

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=19, p=0.3)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

And we want to find this probability:

P(X \geq 5)

And we can use the complement rule:

P(X \geq 5)=1-P(X

We can find the individual probabilities:

P(X=0)=(19C0)(0.3)^0 (1-0.3)^{19-0}=0.00114

P(X=1)=(19C1)(0.3)^1 (1-0.3)^{19-1}=0.0092

P(X=2)=(19C2)(0.3)^2 (1-0.3)^{19-2}=0.0358

P(X=3)=(19C3)(0.3)^3 (1-0.3)^{19-3}=0.0869

P(X=4)=(19C4)(0.3)^4 (1-0.3)^{19-4}=0.1491

And replacing we got:

P(X \geq 5) = 1-[0.00114+0.009282+0.0358+0.0869+0.149]= 0.7178

4 0
3 years ago
What is the least common multiple of 4, 27, 12.
ryzh [129]
The lease common multiple of the set of numbers is 108.

108/4=27.
108/27=4.
108/12=9.
7 0
3 years ago
ed ReviewA company makes bars of soap.soapThe is mixed in large vats, then poured intomolds to make 560 bars of soap. Themixture
inysia [295]
Given:
weight of mixture : 1792 ounces
output of mixture : 560 bars of soap

to get the weight per unit, we simply divide by total weight by the total number of output.

1792 oz / 560 bars = 3.2 ounces per bar.
6 0
3 years ago
Mr. Carson wants to give three pencils to each of his students. There are 26 students in Mr. Carsons class. Boxes of pencils eac
Rasek [7]
Since there are 26 students in Mr. Carson's class, and each kid gets 3 pencils, that means we will multiply 26 and 3, in which we get 78. Now, since boxes of pencils only contain a dozen, which is 12 pencils, we will divide 78 by twelve, in which we get 6.5 boxes. You can't buy half a box, so he will need 7 boxes, since having 6 boxes means that you don't have enough pencils.

Hope this helps!
3 0
3 years ago
Read 2 more answers
Suppose the object moving is Dave, who hasamassofmo =66kgatrest. Whatis Daveâs mass at 90% of the speed of light? At 99% of the
nignag [31]

Step-by-step explanation:

Formula that relates the mass of an object at rest and its mass when it is moving at a speed v:

m=\frac{m_o}{\sqrt{1-\frac{v^2}{c^2}}}

Where :

m = mass of the oebjct in motion

m_o = mass of the object when at rest

v = velocity of a moving object

c = speed of the light = 3\times 0^8 m/s

We have :

1) Mass of the Dave = m_o=66 kg

Velocity of Dave ,v= 90% of speed of light = 0.90c

Mass of the Dave when moving at 90% of the speed of light:

m=\frac{66 kg}{\sqrt{1-\frac{(0.90c)^2}{c^2}}}

m = 151.41 kg

Mass of Dave when when moving at 90% of the speed of light is 151.41 kg.

2) Velocity of Dave ,v= 99% of speed of light = 0.99c

Mass of the Dave when moving at 99% of the speed of light:

m=\frac{66 kg}{\sqrt{1-\frac{(0.99c)^2}{c^2}}}

m = 467.86 kg

Mass of Dave when when moving at 99% of the speed of light is 467.86 kg.

3) Velocity of Dave ,v= 99.9% of speed of light = 0.999c

Mass of the Dave when moving at 99.9% of the speed of light:

m=\frac{66 kg}{\sqrt{1-\frac{(0.999c)^2}{c^2}}}

m = 1,467.17 kg

Mass of Dave when when moving at 99.9% of the speed of light is 1,467.17 kg.

4) Mass of the Dave = m_o=66 kg

Velocity of Dave,v=?

Mass of the Dave when moving at v speed of light: 500

500 kg=\frac{66 kg}{\sqrt{1-\frac{(v)^2}{(3\times 10^8 m/s)^2}}}

v=2.973\times 10^8 m/s

Dave should be moving at speed of 2.973\times 10^8 m/s.

4 0
3 years ago
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