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zepelin [54]
1 year ago
10

It’s supposed to answer in simplest formIf I die is rolled one time find the probability of

Mathematics
1 answer:
Usimov [2.4K]1 year ago
7 0

A die can have 6 possible outcomes.

The probability of an event is calculated using the formula:

P=\frac{number\text{ }of\text{ }required\text{ }outcomes}{number\text{ }of\text{ }total\text{ }outcomes}

Therefore, the probability of rolling a 1 is gotten to be:

P=\frac{1}{6}

The probability is 1/6.

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Find the lcm of 2,3,7 and tell me how u got ur answer
Jobisdone [24]

Answer:

1 is the LCM

Step-by-step explanation:

the factors of 2 = 2 and 1

the factors of 3 = 3 and 1

the factors of 7 = 7 and 1

none of the factors are similar, except for 1 - therefore the LCM is 1

4 0
4 years ago
Which inequality is equivalent to <br> −4t≥28<br> ?
Morgarella [4.7K]

Answer:

t ≤ -7

Step-by-step explanation:

−4t≥28

−4/4 * t ≥ 28/4

−1 * t ≥ 7

-t ≥ 7

t ≤ -7

6 0
4 years ago
You are planning to take to a trip to Montreal, Canada during the month of April and you want to bring clothing that is appropri
EastWind [94]

Answer:

1. E(Y) = 50.54°F

2. SD(Y) = 11.34°F

Step-by-step explanation:

We are given that The daily high temperature X in degrees Celsius in Montreal during April has expected value E(X) = 10.3°C with a standard deviation SD(X) = 3.5°C.

The conversion of X into degrees Fahrenheit Y is Y = (9/5)X + 32.

(1) Y = (9/5)X + 32

   E(Y) = E((9/5)X + 32) = E((9/5)X) + E(32)

           = (9/5) * E(X) + 32   {\because expectation of constant is constant}

           = (9/5) * 10.3 + 32 = 50.54

Therefore, E(Y), the expected daily high in Montreal during April in degrees Fahrenheit is 50.54°F .

(2) Y = (9/5)X + 32

    SD(Y) = SD((9/5)X + 32) = SD((9/5)X) + SD(32)

              = (9/5)^{2} * SD(X) + 0  {\because standard deviation of constant is zero}

              = \beacuse (9/5)^{2} * 3.5 = 11.34°F

Therefore, SD(Y), the standard deviation of the daily high temperature in Montreal during April in degrees Fahrenheit is 11.34°F .

6 0
3 years ago
What is the equation of the line that is perpendicular to the given line and passes through the point (3, 0)?
olga2289 [7]

Step 1

<u>Find the slope of the given line</u>

Let

A(-3,2)\ B(2,-1)

slope mAB is equal to

mAB=\frac{(y2-y1)}{(x2-x1)} \\ \\ mAB=\frac{(-1-2)}{(2+3)} \\ \\ mAB=-\frac{3}{5}

Step 2

<u>Find the slope of the line that is perpendicular to the given line</u>

Let

CD ------> the line that is perpendicular to the given line

we know that

If two lines are perpendicular, then the product of their slopes is equal to -1

so

mAB*mCD=-1\\ mAB=-\frac{3}{5} \\ mCD=-\frac{1}{mAB} \\ mCD=\frac{5}{3}

Step 3

<u>Find the equation of the line with mCD and the point (3,0)</u>

we know that

the equation of the line in the form point-slope is equal to

y-y1=m(x-x1)\\\\ y-0=\frac{5}{3} *(x-3)\\\\ y=\frac{5}{3} x-5

Multiply by 3 both sides

3y=5x-15

5x-3y=15

therefore

the answer is

the equation of the line that is perpendicular to the given line is the equation 5x-3y=15

4 0
3 years ago
Read 2 more answers
Given the function: j(x)=x^2+3x-4 ... Match the description with the critical point.
Assoli18 [71]

Answer:

1

Step-by-step explanation:

3 0
3 years ago
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