In the expression 2(10x) + 15y, x represents the number of hours Nina works in a week at a clothing store and y represents the n

Response:

Step-by-step explanation:

Information given:

Objective:

Calculate the maximum and minimum values

The max value is found as follows:

The min value is identified as follows:

Past due amount = $87.50
Late fee equals 15% of $87.50, which is 0.15 × 87.50 = $13.125
New charges are $75.00

Total due now = 87.50 + 13.125 + 75 = $175.625

Final total: $175.63

The average is also known as the mean, calculated as the sum of values divided by the total number of data points.

Values are h, j, and k

The total data count is 3

The sum of the values is represented as h + j + k

Thus, the mean is [h + j + k] / 3

=> Answer: option c. [h+j+k] over 3

Response:

a. 0.76

b. 0.23

c. 0.5

d. p(B/A) signifies the likelihood that a student with a visa card also possesses a MasterCard.

p(A/B) indicates the probability that a student with a MasterCard also has a visa card.

e. 0.35

f. 0.31

Detailed explanation:

a. p(AUBUC) = P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + P(AnBnC)

        = 0.6 + 0.4 + 0.2 - 0.3 - 0.11 - 0.1 + 0.07 = 0.76

b. P(AnBnC') = P(AnB) - P(AnBnC)

        = 0.3 - 0.07 = 0.23

c. P(B/A) = P(AnB)/P(A)

        = 0.3/0.6 = 0.5

e. P((AnB)/C) = P((AnB)nC)/P(C)

        = P(AnBnC)/P(C)

        = 0.07/0.2 = 0.35

f. P((AUB)/C) = P((AUB)nC)/P(C)

        = (P(AnC) U P(BnC))/P(C)

        = (0.11 + 0.1)/0.2

        = 0.21/0.2 = 0.31

Answer:

Alright, we can express the baseball's motion with an equation like:

h(x) = a*x^2 + b*x + c

Here, x denotes time, while h(x) indicates height.

Let’s construct this:

The acceleration is:

a(t) = a

For velocity, integrating over time results in:

v(x) = a*x + v0

Where v0 signifies the initial vertical velocity.

Subsequently, we can determine position or height by integrating once more:

h(x) = a*x^2 + v0*x + h0

Here, h0 is the initial height.

<pthus our="" equation="" is:="">

h(x) = a*x^2 + v0*x + h0.

<pexamining the="" table:="">

When x = 0s, h(0s) = 6ft

<pthus:>

h(0s) = a*0s^2 + v0*0s + h0 = 6ft

            h0 = 6ft.

It’s also noted that:

h(2s) = h(4s)

<pthe symmetry="" of="" the="" quadratic="" function="" implies="" that="" axis="" lies="" between="" and="" located="" at="" x="3s.&lt;/p"><pin a="" standard="" quadratic="" function:="">

a*x^2 + b*x + c

The symmetry line is given by:

x = -b/2a

<pin this="" instance:="">

b = v0

a = a

<ptherefore we="" derive:="">

3s  = -v0/(2*a)

v0 = -3s*(2a)

<phaving gathered="" all="" necessary="" data="" for="" our="" equation="" we="" can="" express="" it="" as:="">

h(x) = a*x^2 - 6s*a*x + 6ft

<pnext focusing="" on="" just="" one="" variable="" we="" know="" that="" at="" x="2s," h=""><pso:>

h(2s) = 22ft = a*(2s)^2 - 6s*a*2s + 6ft

<pthus our="" resulting="" equation="" reads:="">

h(x) = (-2ft/s^2)*x^2 + (12ft/s)*x + 6ft

b) The height after 5 seconds is expressed as:

h(5s) =  (-2ft/s^2)*(5s)^2 + (12ft/s)*5s + 6ft = 16ft

</pthus></p></pso:></pnext></phaving></ptherefore></pin></pin></pthe></pthus:></pexamining></pthus>