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3 years ago

10

At a concession stand, a pickle and two bags of chips costs a total of $3.25. Three pickles and four bags of chips costs a total

of $7.25. To determine the cost of one pickle, Kevin is writing a system of equations. His first equation is shown below. Which of the following could be the second equation in his system of equations?

1 answer:

kolbaska11 [484]3 years ago

4 0

Answer:

A bag of chips costs $1

A pickle costs $1.25

Step-by-step explanation:

P + 2c = 3.25 Start with these two equations

3p + 4c = 7.25

p = -2c + 3.25 Solve for one variable

3(-2c +3.25) + 4c = 7.25 Substitute

-6c + 9.75 + 4c = 7.25

-2c = -2

c = 1

p + 2(1) = 3.25 Substitute

p + 2 = 3.25

p = 1.25

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A person's blood glucose level and diabetes are closely related. Let x be a random variable measured in milligrams of glucose pe

soldier1979 [14.2K]

Using the normal distribution, it is found that:

a) 0.8599 = 85.99% probability that x is more than 60.

b) 0.1788 = 17.88% probability that x is less than 110.

c) 0.6811 = 68.11% probability that x is between 60 and 110.

d) 0.0643 = 6.43% probability that x is greater than 125.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

The mean is of 87, thus $\mu = 87$ .
The standard deviation is of 25, thus $\sigma = 25$ .

Item a:

This probability is <u>1 subtracted by the p-value of Z when X = 60</u>, thus:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 87}{25}$

$Z = -1.08$

$Z = -1.08$ has a p-value of 0.1401.

1 - 0.1401 = 0.8599

0.8599 = 85.99% probability that x is more than 60.

Item b:

This probability is the <u>p-value of Z when X = 110</u>, thus:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{110 - 87}{25}$

$Z = 0.92$

$Z = 0.92$ has a p-value of 0.8212.

1 - 0.8212 = 0.1788.

0.1788 = 17.88% probability that x is less than 110.

Item c:

This probability is the <u>p-value of Z when X = 110 subtracted by the p-value of Z when X = 60</u>.

From the previous two items, 0.8212 - 0.1401 = 0.6811.

0.6811 = 68.11% probability that x is between 60 and 110.

Item d:

This probability is <u>1 subtracted by the p-value of Z when X = 125</u>, thus:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{125 - 87}{25}$

$Z = 1.52$

$Z = 1.52$ has a p-value of 0.9357.

1 - 0.9357 = 0.0643.

0.0643 = 6.43% probability that x is greater than 125.

A similar problem is given at brainly.com/question/24863330

7 0

3 years ago

What is negative 48 plus 14

leva [86]

Answer: negative 34

Step-by-step explanation:

just take away 14 to 48 and put a negative infront

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3 years ago

What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

as

$\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81$

$\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x$

$\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2$

$\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3$

$\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4$

so equation becomes

$=81+108x+54x^2+12x^3+x^4$

$=x^4+12x^3+54x^2+108x+81$

Therefore,

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

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3 years ago

Which two numbers multiply to -52 and add to 24?

Shtirlitz [24]

Answer: -2 and 26

Step-by-step explanation:

-2 * 26 = -52

-2 + 26 = 24

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marusya05 [52]

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