The formula K=C+273.15 converts temperatures from Celsius C to Kelvin K.

halona walks 1.93 kilometers in 23 minutes, assuming constant speed write a proportion that represents how many kilometers in x

Afina-wow [57]

Dividing 1.93km by 23 minutes, you get 0.083 km per minute, or 5.03 km per hour

4 0

3 years ago

3. From the table below, find Prof. Xin expected value of lateness. (5 points) Lateness P(Lateness) On Time 4/5 1 Hour Late 1/10

wariber [46]

Answer:

The expected value of lateness $\frac{7}{20}$ hours.

Step-by-step explanation:

The probability distribution of lateness is as follows:

Lateness P (Lateness)

On Time 4/5

1 Hour Late 1/10

2 Hours Late 1/20

3 Hours Late 1/20

The formula of expected value of a random variable is:

$E(X)=\sum x\cdot P(X=x)$

Compute the expected value of lateness as follows:

$E(X)=\sum x\cdot P(X=x)$

$=(0\times \frac{4}{5})+(1\times \frac{1}{10})+(2\times \frac{1}{20})+(3\times \frac{1}{20})\\\\=0+\frac{1}{10}+\frac{1}{10}+\frac{3}{20}\\\\=\frac{2+2+3}{20}\\\\=\frac{7}{20}$

Thus, the expected value of lateness $\frac{7}{20}$ hours.

8 0

3 years ago

The region bounded by y=x^2+1, y=x, x=-1, x=2 with square cross sections perpendicular to the x-axis.

VLD [36.1K]

Answer:

The bounded area is 5 + 5/6 square units. (or 35/6 square units)

Step-by-step explanation:

Suppose we want to find the area bounded by two functions f(x) and g(x) in a given interval (x1, x2)

Such that f(x) > g(x) in the given interval.

This area then can be calculated as the integral between x1 and x2 for f(x) - g(x).

We want to find the area bounded by:

f(x) = y = x^2 + 1

g(x) = y = x

x = -1

x = 2

To find this area, we need to f(x) - g(x) between x = -1 and x = 2

This is:

$\int\limits^2_{-1} {(f(x) - g(x))} \, dx$

$\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx$

We know that:

$\int\limits^{}_{} {x} \, dx = \frac{x^2}{2}$

$\int\limits^{}_{} {1} \, dx = x$

$\int\limits^{}_{} {x^2} \, dx = \frac{x^3}{3}$

Then our integral is:

$\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx = (\frac{2^3}{2} + 2 - \frac{2^2}{2}) - (\frac{(-1)^3}{3} + (-1) - \frac{(-1)^2}{2} )$

The right side is equal to:

$(4 + 2 - 2) - ( -1/3 - 1 - 1/2) = 4 + 1/3 + 1 + 1/2 = 5 + 2/6 + 3/6 = 5 + 5/6$

The bounded area is 5 + 5/6 square units.

3 0

3 years ago

Name the line of reflection used to map each preimage to its image

rewona [7]

Answer:

From the picture attached,

9). When pre-image KLMN is a reflected across x = -2, we get image K'L'M'N'.

Therefore, line of reflection will be x = -2

10). In this picture line of reflection which passes through origin (0, 0) and slope = 1 will be,

y = x.

11). Since, all the points of image and preimage points are equidistant from x-axis, line of reflection will be x-axis or y = 0.

12). Line passing through midpoint of the line joining BB' will be the line of reflection.

Coordinates of B → (5, 3)

Coordinates of B' → (5, 7)

Coordinates of BB' → → (5, 5)

Therefore, line of reflection will be y = 5.

13). Coordinates of G → (2, 5)

Coordinates of G' → (-2, 5)

Midpoint joining the segment GG' → → (0, 5)

Therefore, line of reflection → x = 0

14). Line of reflection passing through (0, 0) and slope = -1

y = -x

Step-by-step explanation:

3 0

2 years ago

The average test scores for a particular test in algebra was 84 with a standard deviation of 5. What percentage of the students

Delicious77 [7]

Answer:

The percentage of the students scored higher than 89% is 15.87%

Step-by-step explanation:

we are given

The average test scores for a particular test in algebra was 84

so, mean is 84

$\mu=84$

a standard deviation of 5

so,

$\sigma=5$

we have

x=89

now, we can use z-score formula

$z=\frac{x-\mu}{\sigma}$

now, we can plug values

$z=\frac{89-84}{5}$

$z=1$

now, we can use calculator

and we get

$p(z>1)=0.1587$

So,

The percentage of the students scored higher than 89% is 15.87%

8 0

4 years ago