Find the sum of: (4.3 x 10¹¹) + (9.2 x 10¹²) *

Pls solve quickly !!

Eva8 [605]

Answer:

Options A

Step-by-step explanation:

Properties of the graph given for the function 'g',

y-intercept of the function 'g' → y = -4

Minimum value of g(x) → y = -4 (Value of the function at the lowest point)

Roots of g(x) → x = -2, 2

At x = 4,

g(4) = 11

Another function is,

f(x) = -x² - 4x - 4

= -(x² + 4x + 4)

= -(x + 2)²

Properties of the function f(x) = -(x + 2)²,

y-intercept (at x = 0) of the function 'f',

y = -(x + 2)²

y = -4

Since, leading coefficient of the function is (-1) therefore, graph will open downwards and the maximum point will be its vertex.

Maximum value of the function 'f' = Value of the function at the vertex

Coordinates of vertex → (-2, 0)

Therefore, maximum value of 'f' = 0

Roots (at y = 0) of the function 'f' → x = -2

At x = 4,

f(4) = -(4 + 2)²

= -36

Therefore, Options A will be the correct option.

5 0

3 years ago

A bowl contains 80 M&M candies. If 15 of the candies are orange, what fraction of the candies is not orange?

lesantik [10]

65 candies are not orange.

5 0

3 years ago

Read 2 more answers

Mr. Good Wrench advertises that a customer will have to wait no more than 30 minutes for an oil change. A sample of 26 oil chang

Andru [333]

Answer:

The 90% confidence interval for the population standard deviation waiting time for an oil change is (3.9, 6.3).

Step-by-step explanation:

The (1 - α)% confidence interval for the population standard deviation is:

$CI=\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2, (n-1)}}}\leq \sigma\leq \sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2, (n-1)}}}$

The information provided is:

n = 26

s = 4.8 minutes

Confidence level = 90%

Compute the critical values of Chi-square as follows:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.10/2, (26-1)}=\chi^{2}_{0.05, 25}=37.652$

$\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.10/2, (26-1)}=\chi^{2}_{0.95, 25}=14.611$

*Use a Chi-square table.

Compute the 90% confidence interval for the population standard deviation waiting time for an oil change as follows:

$CI=\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2, (n-1)}}}\leq \sigma\leq \sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2, (n-1)}}}$

$=\sqrt{\frac{(26-1)\times 4.8^{2}}{37.652}}\leq \sigma\leq \sqrt{\frac{(26-1)\times 4.8^{2}}{14.611}}\\\\=3.9113\leq \sigma\leq 6.2787\\\\\approx 3.9 \leq \sigma\leq6.3$

Thus, the 90% confidence interval for the population standard deviation waiting time for an oil change is (3.9, 6.3).

3 0

3 years ago

suppose that the water level of a river is 34 feet and that it is receding at a rate of 0.5 foot per day. Write an equation for

lubasha [3.4K]

Equation: L= -0.5D+34.

The water level be 26 feet in 16 days.

Step-by-step explanation:

Let us consider the initial level (x=0) of the water is 34 feet. Then its coordinate point can be written as (0,34).

The water is receding at the rate 0.5 feet per day, which given us the information that 0.5 is the slope since it is the rate of change and it is decreasing so it is a negative slope.

⇒m= -0.5.

The line equation is y = mx+c. Let us rewrite as L= mD+c.

Where L is the level of water(y-axis) and D is the days(x-axis) and c is y-intercept.

⇒L= (-0.5)D+c.

To find y-intercept, substitute (0,34) (the known point from the graph) in the equation. Also, the point in y axis when x=0 is y-intercept.

34= (-0.5)(0)+c.

c=34.

∴L= (-0.5)D+34.

To find the days when the water achieves level as 26 feet:

26 = (-0.5)D+34.

26-34= (-0.5)D.

-8 = (-0.5)D.

D= $\frac{-8}{-0.5}$ .

D= 16.

7 0

4 years ago

The rectangular floor of a classroom is 32 feet in length and 36 feet in width. A scale drawing of the floor has a length of 8 i

Vanyuwa [196]

Answer:

34

Step-by-step explanation:

32 / 8 = 4
36 / 4 = 9

perimeter of rectangle = 2(Length) + 2(width)
P = 2(8) + 2(9)
P = 16 + 18
P = 34

6 0

2 years ago