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4 years ago

The sum of two numbers is -316. One number is 94 less than the other. Find the numbers.

Mathematics

2 answers:

Oksanka [162]4 years ago

5 0

Answer:

-94+-316=-410

-410+94=-316

Step-by-step explanation:

shutvik [7]4 years ago

5 0

Answer:

One number is -111 and the other is -205.

Step-by-step explanation:

First, form the equation. Since the two numbers equal -316, the equation would equal -316. The equation is as follows:

-316 = n + (n - 94).

To solve you do the following steps:

1. Remove the parenthesis. The new equation is -316 = n + n - 94

2. Combine like terms: -316 = 2n - 94

3. Move the 2n to the left side of the equation by subtracting it from both sides:

-316 = 2n - 94

-2n -2n

-2n - 316 = -94

4. Move the -316 to the right side of the equation by adding it to both sides:

-2n - 316 = -9

+ 316 + 316

-2n = 222

5. Divide both sides by - 2

-2n/ -2 = n

222/-2 = -111

Therefore, n = -111

To get the other number, subtract 94 from -111.

-111 - 94 = -205.

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A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

EleoNora [17]

Answer:

a) $P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

b) $P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

c) $P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

d) $E(X) = 20*0.2= 4$

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:

$X \sim Binom(n=20, p=0.2)$

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:

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

Part a

We want this probability:

$P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)$

And we can use the probability mass function and we got:

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

And adding we got:

$P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061$

Part b

We want this probability:

$P(X=4)$

And using the probability mass function we got:

$P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182$

Part c

We want this probability:

$P(X>3)$

We can use the complement rule and we got:

$P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]$

$P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115$

$P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576$

$P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369$

$P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054$

And replacing we got:

$P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886$

Part d

The expected value is given by:

$E(X) = np$

And replacing we got:

$E(X) = 20*0.2= 4$

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3 years ago

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Answer:

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Discount = 25% of MP = m/4

SP = MP - Discount

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If profit is 20%, CP = 3m/4 x 100/120 = 5m/8

=> Profit = 3m/4 - 5m/8 = m/8

Given Profit - Discount = Rs 15

=> |m/8 - m/4| = 15

m/8 = 15

m = Rs 120

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3 years ago

Given: ABCD is a rhombus. Prove: △AEB is a right triangle.

lukranit [14]

Answer:

Step-by-step explanation:

∵ ABCD is a rhombus

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∴ ΔAEB is right angled triangle

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3 years ago

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s344n2d4d5 [400]

I hope this helps you

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3 years ago

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Answer:

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3 years ago