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2 years ago

Diana is going to roll a 6-sided die. What's the probability she will role a 4 or a 5? *

Mathematics

1 answer:

anastassius [24]2 years ago

6 0

The probability of rolling any one number on a 6 sided die is 1/6

The probability of rolling a 4 is 1/6

The probability of rolling a 5 is 1/6

The probability of rolling a 4 or a 5 is the probability of rolling a 4 plus the probability of rolling a 5:

1/6 + 1/6 = 2/6 = 1/3

The answer is 1/3

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What is the stem and leaf plot

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3 years ago

Graph the line with slope 7 and y-intercept -7

poizon [28]

Given :-

To graph the line with slope 7 and y intercept -7.

To Find :-

The graph of the line .

Solution :-

Here since the slope of the line is 7 and y intercept is -7 , we can use the slope intercept form of the line to find the equation of the line . The slope intercept form of the line is ,

$\longrightarrow$ y = mx + c

On putting the respective values ,

$\longrightarrow$ y = 7(x) + (-7)

$\longrightarrow$ y = 7x - 7

For the graph see attachment .

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1 year ago

Marta conducts emissions inspections on cars. She finds that 8 percent of the cars fail the inspection. Let X be the number of c

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Answer:

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Step-by-step explanation:

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3 years ago

Express in simplest form -45%

Ksenya-84 [330]

Answer: hello, your answer is -0.45

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2 years ago

PLEASE HELP!!!!!!!dgbdgdbhdndcn

bogdanovich [222]

Problem 1)

AC is only perpendicular to EF if angle ADE is 90 degrees

(angle ADE) + (angle DAE) + (angle AED) = 180
(angle ADE) + (44) + (48) = 180
(angle ADE) + 92 = 180
(angle ADE) + 92 - 92 = 180 - 92
angle ADE = 88

Since angle ADE is actually 88 degrees, we do NOT have a right angle so we do NOT have a right triangle

Triangle AED is acute (all 3 angles are less than 90 degrees)

So because angle ADE is NOT 90 degrees, this means AC is NOT perpendicular to EF

-------------------------------------------------------------

Problem 2)

a) The center is (2,-3)

The center is (h,k) and we can see that h = 2 and k = -3. It might help to write (x-2)^2+(y+3)^2 = 9 into (x-2)^2+(y-(-3))^2 = 3^3 then compare it to (x-h)^2 + (y-k)^2 = r^2

---------------------

b) The radius is 3 and the diameter is 6

From part a), we have (x-2)^2+(y-(-3))^2 = 3^3 matching (x-h)^2 + (y-k)^2 = r^2

where
h = 2
k = -3
r = 3

so, radius = r = 3
diameter = d = 2*r = 2*3 = 6

---------------------

c) The graph is shown in the image attachment. It is a circle with center point C = (2,-3) and radius r = 3.

Some points on the circle are

A = (2, 0)
B = (5, -3)
D = (2, -6)
E = (-1, -3)

Note how the distance from the center C to some point on the circle, say point B, is 3 units. In other words segment BC = 3.

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3 years ago