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Afina-wow [57]
3 years ago
9

The mode, the range and the mean of the following set of data are, respectively

Mathematics
1 answer:
alina1380 [7]3 years ago
5 0

Answer:

yes

Step-by-step explanation:

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Prove this identity sin(2A)=2sinAcosA.
34kurt
The solution for proving the identity is as follows:

sin(2A) = sin(A + A) 
As sin(a + b) = sinacosb + sinbcosa, 
<span>sin(A + A) = sinAcosA + sinAcosA 
</span>
<span>Therefore, sin(2A) = 2sinAcosA

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your inquiries and questions soon. Have a nice day ahead!
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7 0
2 years ago
The length of a rectangle is 8cm and the height is 16cm find the answer of the above figures using this formular...
notsponge [240]
  • Length=8cm
  • Breadth=16cm(Height)

<h3>Perimeter:-</h3>

\\ \sf\longmapsto Perimeter=2(L+B)

\\ \sf\longmapsto Perimeter=2(16+8)

\\ \sf\longmapsto Perimeter=2(24)

\\ \sf\longmapsto Perimeter=48cm

Area:-

\\ \sf\longmapsto Area=Length\times Breadth

\\ \sf\longmapsto Area=8(16)

\\ \sf\longmapsto Area=128cm^2

3 0
2 years ago
Y=x+5<br> 4x+y=20<br> Solving system by substitution
sineoko [7]
Y = x + 5....so we sub in x + 5 for y in the other equation

4x + y = 20
4x + x + 5 = 20
5x + 5 = 20
5x = 20 - 5
5x = 15
x = 15/5
x = 3

y = x + 5
y = 3 + 5
y = 8

solution is (3,8)
3 0
3 years ago
Please help me asap!!
Amiraneli [1.4K]

Answer:

D

Step-by-step explanation:

6 0
3 years ago
You are at a stall at a fair where you have to throw a ball at a target. There are two versions of the game. In the first
Tomtit [17]

Answer:

P(X=0)=(3C0)(0.1)^0 (1-0.1)^{3-0}=0.729

And the probability of loss with the first wersion is 0.729

P(Y=0)=(5C0)(0.05)^0 (1-0.05)^{5-0}=0.774

And the probability of loss with the first wersion is 0.774

As we can see the best alternative is the first version since the probability of loss is lower than the probability of loss on version 2.

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

Alternative 1

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

X \sim Binom(n=3, p=0.1)

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!}

We can find the probability of loss like this P(X=0) and if we find this probability we got this:

P(X=0)=(3C0)(0.1)^0 (1-0.1)^{3-0}=0.729

And the probability of loss with the first wersion is 0.729

Alternative 2

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

Y \sim Binom(n=5, p=0.05)

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

P(Y)=(nCy)(p)^y (1-p)^{n-y}

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

nCy=\frac{n!}{(n-y)! y!}

We can find the probability of loss like this P(Y=0) and if we find this probability we got this:

P(Y=0)=(5C0)(0.05)^0 (1-0.05)^{5-0}=0.774

And the probability of loss with the first wersion is 0.774

As we can see the best alternative is the first version since the probability of loss is lower than the probability of loss on version 2.

4 0
3 years ago
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