Eight times the sum of 5 and a number is less than 56.

Company A charges a $80 annual fee plus a $10/hr car share fee. Company B

icang [17]

Answer:

The number of hours must be greater than 9 to make a Company B a better deal

Step-by-step explanation:

Let

x ----> the number of hours

y ----> the total charge in dollars

we know that

The linear equation in slope intercept form is equal to

$y=mx+b$

where

m is the slope or unit rate of the linear equation

b is the y-intercept or initial value of the linear equation

In this problem we have

Company A

The slope is equal to $m=\$10\ per\ hour$

The y-intercept is equal to $b=\$80$

so

The linear equation is equal to

$y=10x+80$ -----> equation A

Company B

The slope is equal to $m=\$5\ per\ hour$

The y-intercept is equal to $b=\$125$

so

The linear equation is equal to

$y=5x+125$ -----> equation B

If a Company B is a better deal

then

$5x+125 < 10x+80$

solve for x

$125-80 < 10x-5x$

$45 < 5x$

$9 < x$

Rewrite

$x> 9\ hours$

therefore

The number of hours must be greater than 9 to make a Company B a better deal

7 0

3 years ago

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Graph y= 1/2x–3 . Please. Just please.

agasfer [191]

I don't have a graph, but I can try to show you how to do it.

Go to -3 on the y-axis (the vertical one in the middle)

Put a dot there. That's your y-intercept.

From the dot go up 1 line, then go right 2 lines.

Put a dot where you end up. Continue doing so until you get enough for your graph.

3 0

3 years ago

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Solve the given initial value problem and determine how the interval in which the solution exists depends on the initial value y

zalisa [80]

Answer:

y has a finite solution for any value y_0 ≠ 0.

Step-by-step explanation:

Given the differential equation

y' + y³ = 0

We can rewrite this as

dy/dx + y³ = 0

Multiplying through by dx

dy + y³dx = 0

Divide through by y³, we have

dy/y³ + dx = 0

dy/y³ = -dx

Integrating both sides

-1/(2y²) = - x + c

Multiplying through by -1, we have

1/(2y²) = x + C (Where C = -c)

Applying the initial condition y(0) = y_0, put x = 0, and y = y_0

1/(2y_0²) = 0 + C

C = 1/(2y_0²)

So

1/(2y²) = x + 1/(2y_0²)

2y² = 1/[x + 1/(2y_0²)]

y² = 1/[2x + 1/(y_0²)]

y = 1/[2x + 1/(y_0²)]½

This is the required solution to the initial value problem.

The interval of the solution depends on the value of y_0. There are infinitely many solutions for y_0 assumes a real number.

For y_0 = 0, the solution has an expression 1/0, which makes the solution infinite.

With this, y has a finite solution for any value y_0 ≠ 0.

8 0

4 years ago

You have a biased coin for which p(h)=pp(h)=p. you toss the coin 2020 times. what is the probability that you observe 88 heads a

Nat2105 [25]

The question describes a binomial probability with p(h) = p, then p(t) = 1 - p and number of trials (n) = 20

The probability of a binomial distribution is given by

$P(x)=\, ^nC_xp^x(1-p)^{n-x}$

Part A:

The probability of observing 8 heads and 12 tails is given by:

$P(8)=\, ^{20}C_8p^8(1-p)^{20-8}=\, ^{20}C_8p^8(1-p)^{12}$

Part B:

You observe more than 8 heads and more than 8 tails, when you observe 9 heads and 11 tails, 10 heads and 10 tails, and 11 heads and 9 tails.

Therefore, the probability of observing more than 8 heads and more than 8 tails is given by:

$P(9)+P(10)+P(11) \\ \\ =\, ^{20}C_9p^9(1-p)^{20-9}+\, ^{20}C_{10}p^{10}(1-p)^{20-10}+\, ^{20}C_{11}p^{11}(1-p)^{20-11} \\ \\ =\, ^{20}C_9p^9(1-p)^{11}+\, ^{20}C_{10}p^{10}(1-p)^{10}+\, ^{20}C_{11}p^{11}(1-p)^{9}$

7 0

3 years ago

What is a reasonable estimate for the product 43 times 68

dsp73

43 times 68.

43
x68

8 times 3 is 24. Bring the 2 over the 4.
8 times 4 is 32. Add 4. Equals 36.
Answer so far is 364.

Add a 0 before answering.
3 times 6 is 18. Bring the 1 over the 4.
4 times 6 is 24. Add 4. Equals 28.
Answer so far is 288

Add 364 + 288
Final answer is 652.

7 0

3 years ago

Read 2 more answers