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2 years ago

7

The table shows some of the values of a function of the form y=ax^2+bx+c which is the value of c, the constant, the constant for

the function?

1 answer:

cupoosta [38]2 years ago

5 0

Answer:

$c=8$

Step-by-step explanation:

Given: $y=ax^2+bx+c$

To find: c

Solution:

At $x=4,y=48$ :

$48=16a+4b+c$ (i)

At $x=5,y=73$ :

$73=25a+5b+c$ (ii)

At $x=6,y=104$ :

$104=36a+6b+c$ (iii)

On subtracting (i) and (ii),

$25=9a+b$ (iv)

On subtracting (i) and (iii),

$56=20a+2b\\\\28=10a+b \,\,\,\,(v)$

On subtracting (iv) and (v),

$a=3$

Put a=3 in (iv)

$25=9(3)+b\\25=27+b\\25-27=b\\-2=b$

Put $a=3,b=-2$ in (iii)

$104=36(3)+6(-2)+c\\104=108-12+c\\104=96+c\\104-96=c\\8=c$

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A rectangle has height and width changing in such a way that the area remains constant 2 square feet. At the instant the height

hichkok12 [17]

Answer:

$\frac{1}{3}$ ft/min is the rate of changing of width

Step-by-step explanation:

Given -

The area always remain constant i.2 2 square feet.

Height of the rectangle = 2 feet

Rate of changing of height = 6 feet per minute

Since area is constant

2 sq ft = (2 * 6) ft/min * 1 * x ft/min

x = $\frac{1}{3}$ ft/min

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2 years ago

A local cable company claims that the proportion of people who have Internet access is less than 63%. To test this claim, a rand

MaRussiya [10]

Answer:

Step-by-step explanation:

For the null hypothesis,

H0 : p = 0.63

For the alternative hypothesis,

Ha : p < 0.63

This is a left tailed test

Considering the population proportion, probability of success, p = 0.63

q = probability of failure = 1 - p

q = 1 - 0.63 = 0.37

Considering the sample,

Sample proportion, P = x/n

Where

x = number of success = 478

n = number of samples = 800

P = 478/800 = 0.6

We would determine the test statistic which is the z score

z = (P - p)/√pq/n

z = (0.6 - 0.63)/√(0.63 × 0.37)/800 = - 1.76

From the normal distribution table, the area below the test z score in the left tail 0.039

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7 0

2 years ago

A supermarket has two customers waiting to pay for their purchases at counter I and one customer waiting to pay at counter II. L

Pachacha [2.7K]

Answer:

b. 0.864

Step-by-step explanation:

Let's start defining the random variables.

Y1 : ''Number of customers who spend more than $50 on groceries at counter 1''

Y2 : ''Number of customers who spend more than $50 on groceries at counter 2''

If X is a binomial random variable, the probability function for X is :

$P(X=x)=(nCx)p^{x}(1-p)^{n-x}$

Where P(X=x) is the probability of the random variable X to assume the value x

nCx is the combinatorial number define as :

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

n is the number of independent Bernoulli experiments taking place

And p is the success probability.

In counter I :

Y1 ~ Bi (n,p)

Y1 ~ Bi(2,0.2)

$P(Y1=y1)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}$

With y1 ∈ {0,1,2}

And P( Y1 = y1 ) = 0 with y1 ∉ {0,1,2}

In counter II :

Y2 ~ Bi (n,p)

Y2 ~ Bi (1,0.3)

$P(Y2=y2)=(1Cy2)(0.3)^{y2}(0.7)^{1-y2}$

With y2 ∈ {0,1}

And P( Y2 = y2 ) = 0 with y2 ∉ {0,1}

(1Cy2) with y2 = 0 and y2 = 1 is equal to 1 so the probability function for Y2 is :

$P(Y2=y2)=(0.3)^{y2}(0.7)^{1-y2}$

Y1 and Y2 are independent so the joint probability distribution is the product of the Y1 probability function and the Y2 probability function.

$P(Y1=y1,Y2=y2)=P(Y1=y1).P(Y2=y2)$

$P(Y1=y1,Y2=y2)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}(0.3)^{y2}(0.7)^{1-y2}$

With y1 ∈ {0,1,2} and y2 ∈ {0,1}

P( Y1 = y1 , Y2 = y2) = 0 when y1 ∉ {0,1,2} or y2 ∉ {0,1}

b. Not more than one of three customers will spend more than $50 can mathematically be expressed as :

$Y1 + Y2 \leq 1$

$Y1 + Y2\leq 1$ when Y1 = 0 and Y2 = 0 , when Y1 = 1 and Y2 = 0 and finally when Y1 = 0 and Y2 = 1

To calculate $P(Y1+Y2\leq 1)$ we must sume all the probabilities that satisfy the equation :

$P(Y1+Y2\leq 1)=P(Y1=0,Y2=0)+P(Y1=1,Y2=0)+P(Y1=0,Y2=1)$

$P(Y1=0,Y2=0)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{0}(0.7)^{1-0}=(0.8)^{2}(0.7)=0.448$

$P(Y1=1,Y2=0)=(2C1)(0.2)^{1}(0.8)^{2-1}(0.3)^{0}(0.7)^{1-0}=2(0.2)(0.8)(0.7)=0.224$

$P(Y1=0,Y2=1)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{1}(0.7)^{1-1}=(0.8)^{2}(0.3)=0.192$

$P(Y1+Y2\leq 1)=0.448+0.224+0.192=0.864\\P(Y1+Y2\leq 1)=0.864$

7 0

2 years ago

Amee received an end of year bonus of $1550 at work and went on a shopping spree. She spent $225 at the department store, $275 a

atroni [7]

Answer: She has $1022 left.

Step-by-step explanation:

The total amount that Amee received an the end of year as bonus at work is $1550.

She went on a shopping spree, spending $225 at the department store, $275 at the home furnishing store, and $28 at the card shop. Therefore, the total amount that she spent at the department store, the home furnishing store, and the card shop is

225 + 275 + 28 = $528

Therefore, the total amount of her bonus that she has left is

1550 - 528 = $1022

6 0

2 years ago

What is the product? StartFraction 3 k Over k + 1 EndFraction times StartFraction k squared minus 1 Over 3 k cubed EndFraction S

Nana76 [90]

Answer:

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Step-by-step explanation:

3k / (k + 1) × (k²- 1) / 3k³

= 3k(k² - 1) / (k + 1)(3k³)

= 3k³ - 3k / 3k⁴ + 3k³

= -3k / 3k⁴

= -1/k³

StartFraction k + 1 Over k squared EndFraction

(k + 1) / k²

StartFraction k minus 1 Over k squared EndFraction

(k - 1)/k²

StartFraction negative 1 Over k cubed EndFraction

= -1/k³

StartFraction 1 Over k EndFraction

= 1/k

4 0

1 year ago