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3 years ago

15

cory bought 4 apples and 6 oranges and paid 5.10. Dalia bought 2 apples and 5 oranges and paid 3.65. what is the cost of 2 orang

es?

1 answer:

aev [14]3 years ago

4 0

Answer:

pls tell me where cory bought them from. its a very cheap price . i want to buy

Step-by-step explanation:

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3 years ago

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Will give Brainlest Answer please answer i need help

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Answer:

C. {-1, 5, 8}

Step-by-step explanation:

Use each of the domain values in the function to see what the corresponding range value is.

f(-1) = 5 -3(-1) = 8

f(0) = 5 -3(0) = 5

f(2) = 5 -3(2) = -1

The range is the set of numbers {-1, 5, 8}.

_____

<em>Additional comment</em>

The values in a set are generally listed lowest to highest. The coefficient of x in the equation for f(x) is negative, meaning the lowest range value will correspond to the highest domain value. If you start by finding f(2) = -1, you immediately eliminate all answer choices except B and C.

Those choices differ only in the middle value, so you can tell which is correct by evaluating f(x) for the middle domain value: f(0) = 5. Only one answer choice has both -1 and 5 in the set.

(There are two answers here: how you work the problem, and how you game a multiple choice question.)

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3 years ago

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

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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:

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

3 0

3 years ago

Add.<br> (-x^3+26x^2-7-13)+(6x^4-x^3+8x+27)<br> express the answer in standard form.

Alex777 [14]

$(-x^3+26x^2-7-13)+(6x^4-x^3+8x+27)\\\\=-x^3+26x^2-7-13+6x^4-x^3+8x+27\\\\=6x^4+(-x^3-x^3)+26x^2+8x+(27-7-13)\\\\=\boxed{6x^4-2x^3+26x^2+8x+7}$

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3 years ago