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MArishka [77]
3 years ago
9

On a cold February morning, the radiator fluid in Stanley’s car is -18 degrees When the engine is running, the temperature goes

up 4.5 degrees per minute. Approximately how long will it take before the radiator fluid temperature reaches 60 degrees ?
Mathematics
1 answer:
Vitek1552 [10]3 years ago
7 0

To figure this out you first find the difference between -18 and 60, which is 78 you then just divide, 78/4.5 which is approximately 17.55 mins or 17.6 rounded.

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Write this decimal in a fraction in its simplest form.
sammy [17]
The simplified form of that decimal in fraction form is 3/20. Hope this helps.
5 0
2 years ago
Hi, can you help me answer this question, please, thank you!
frutty [35]

Given:

mean, u = 0

standard deviation σ = 1

Let's determine the following:

(a) Probability of an outcome that is more than -1.26.

Here, we are to find: P(x > -1.26).

Apply the formula:

z=\frac{x^{\prime}-u}{\sigma}

Thus, we have:

\begin{gathered} P(x>-1.26)=\frac{-1.26-0}{1} \\  \\ P(x>-1.26)=\frac{-1.26}{1}=-1.26 \end{gathered}

Using the standard normal table, we have:

NORMSDIST(-1.26) = 0.1038

Therefore, the probability of an outcome that is more than -1.26 is 0.1038

(b) Probability of an outcome that

4 0
11 months ago
I have 100 items of product in stock. The probability mass function for the product's demand D is P(D=90)=P(D=100)=P(D=110)=1/3.
masya89 [10]

Answer:

The probability mass function for the items sold is

P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.

The mean is 96.667

The variance is 22.222

b) The probability mass function for the unfilled demand due to lack of stock is

P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.

The mean is 3.333

The variance is 33.333

Step-by-step explanation:

If the demand is higher than 100, then you will sell 100 items only. Thus, there is a probability of 1/3+1/3 = 2/3 that you will sell 100 items, while there is a probability of 1/3 that you will sell 90.

The probability mass function for the items sold is

P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.

The mean is 1/3 * 90 + 2/3 * 100 = 290/3 = 96.667

The variance is V(X) = E(X²)-E(X)² = (1/3*90² + 2/3*100²) - (290/3)² = 200/9 = 22.222

b) If order to be unfilled demand, you need to have a demand of 110, which happens with probability 1/3. In that case, the value of the variable, lets call it Y, that counts the amount of unfilled demand due to lack of stock is 110-100 = 10. In any other case, the value of Y is 0, which would happen with probability 1-1/3 = 2/3. Thus

P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.

The mean is 2/3 * 0 + 1/3 * 10 = 10/3 = 3.333

The variance is 2/3*0² + 1/3*10² = 100/3 = 33.333

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2 years ago
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Answer:

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Answer:

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Step-by-step explanation:

i think answer its c

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