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exis [7]
3 years ago
13

Brand X costs $2.70 for 10 ounces. Brand Y costs 4 cents more per ounces. What is the cost of 11 ounces of brand Y?

Mathematics
1 answer:
Ksju [112]3 years ago
8 0
Well you have to first find the constant of Brand X which would be 2.70/10 = .27. Brand Y is .4 more so that would be .31 per ounce. So then .31 x 11 = 3.41. The cost of 11 ounces of brand Y is $3.41.
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TEKS 7.10.A, 7.10.C
ch4aika [34]

Answer:

a=possible amount

3a+28>200

Step-by-step explanation:

3 0
3 years ago
And a triangle.
luda_lava [24]

Answer:

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3 0
3 years ago
In a particular faculty 60% of students are men and 40% are women. In a random sample of 50 students what is the probability tha
zimovet [89]

Answer:

a) The expected value is given by:

E(X) = np = 50*0.4 = 20

and the variance is given by:

Var(X) =np(1-p) = 50*0.4*(1-0.4) = 12

b) P(X>25)= 1-P(X\leq 25)

And we can find this probability with the following Excel code:

=1-BINOM.DIST(25,50,0.4,TRUE)

And we got:

P(X>25)= 1-P(X\leq 25)=0.0573

c) 1) Random sample (assumed)

2) np= 50*0.4= 20 >10

n(1-p) =50*0.6= 30>10

3) Independence (assumed)

Since the 3 conditions are satisfied we can use the normal approximation:

X \sim N(\mu = 20 , \sigma= 3.464)

d) P(X>25) = 1-P(Z< \frac{25-20}{3.464}) = 1-P(z

e) P(X>25)= P(X>25.5) = 1-P(X \leq 25.5)

P(X>25)= P(X>25.5) = 1-P(X \leq 25.5)= 1-P(Z

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".

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

X \sim Binom(n=50, p=0.4)

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

The expected value is given by:

E(X) = np = 50*0.4 = 20

and the variance is given by:

Var(X) =np(1-p) = 50*0.4*(1-0.4) = 12

Part b

For this case we want to find this probability:

P(X>25)= 1-P(X\leq 25)

And we can find this probability with the following Excel code:

=1-BINOM.DIST(25,50,0.4,TRUE)

And we got:

P(X>25)= 1-P(X\leq 25)=0.0573

Part c

1) Random sample (assumed)

2) np= 50*0.4= 20 >10

n(1-p) =50*0.6= 30>10

3) Independence (assumed)

Since the 3 conditions are satisfied we can use the normal approximation:

X \sim N(\mu = 20 , \sigma= 3.464)

Part d

We want this probability:

P(X>25) = 1-P(Z< \frac{25-20}{3.464}) = 1-P(z

Part e

For this case we use the continuity correction and we have this:

P(X>25)= P(X>25.5) = 1-P(X \leq 25.5)

P(X>25)= P(X>25.5) = 1-P(X \leq 25.5)= 1-P(Z

4 0
3 years ago
Help ASAp 3mins left
IRINA_888 [86]
You didn’t attach anything?..
4 0
3 years ago
Read 2 more answers
F(x) = 5x2 − 70x + 258. Complete the square.
astraxan [27]
5(x^2 - 14x + 258/5)
= 5((x-7)^2 -49+258/5)
= 5((x-7)^2 + 13/5)
= 5(x-7)^2 + 13
I belive this is how it should be done
7 0
3 years ago
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