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const2013 [10]
3 years ago
15

Well can someone help get the answer in at least 4 minutes tops

Mathematics
2 answers:
erik [133]3 years ago
8 0
Answer: Yup its C, 14 I can confirm this.
Alex3 years ago
6 0

Answer: 14 is the answer, C.

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you and 3 friends pay $26.55 for a pizza and 4 of the same kind of drink. The pizza cost $18.75. write and solve an equation tha
Anika [276]
Total-pizza/4 drinks= price of one drink

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3 years ago
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3 years ago
A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday�s mail. In actuali
Softa [21]

Answer:

P(Y = 0) = 0.09

P(Y = 1) = 0.4

P(Y = 2) = 0.32

P(Y = 3) = 0.19

Step-by-step explanation:

Let the events be:

W = Wednesday

T = Thursday

F = Friday

S = Saturday

Their corresponding probabilities are

P(W) = 0.3\\P(T) = 0.4\\P(F) = 0.2\\P(S) = 0.1

Since Y = number of days beyond Wednesday that it takes for both magazines to arrive(so possible Y values are 0, 1, 2 or 3)

The possible number of outcomes are therefore 4^2 = 16\\(W, W), (W, T), (W, F), (W, S)\\(T, W), (T, T), (T, F), (T, S)\\(F, W), (F, T), (F, F), (F, S)\\(S, W), (S, T), (S, F), (S, S)

The values associated for each of the outcomes are as follows:

Y(W, W) = 0, Y(W, T) = 1, Y(W, F) = 2, Y(W, S) = 3\\Y(T, W) = 1, Y(T, T) = 1, Y(T, F) = 2, Y(T, S) = 3\\Y(F, W) = 2, Y(F, T) = 2, Y(F, F) = 2, Y(F, S) = 3\\Y(S, W) = 3, Y(S, T) = 3, Y(S, F) = 3, Y(S, S) = 3

The probability mass function of Y is,

P(Y = 0) = 0.3(0.3) = 0.09\\P(Y = 1) = P[(W, T) or (T, W) or (T, T)]\\= [0.3(0.4) + 0.3(0.4) + 0.4(0.4)]\\= 0.4\\\\P(Y = 2) = P[(W, F) or (T, F) or (F, W) or (F, T) or (F, F)]\\= [0.3(0.2) + 0.4(0.2) + 0.2(0.3) + 0.2(0.4) + 0.2(0.2)]\\= 0.32\\\\P(Y = 3) = P[(W, S) or (T, S) or (F, S) or (S, W) or (S, T) or (S, F) or (S, S)]\\= [0.3(0.1) + 0.4(0.1) + 0.2(0.1) + 0.1(0.3) 0.1(0.4) + 0.1(0.2) + 0.1(0.1)]\\= 0.19

7 0
3 years ago
Find the side in of point a to point c
vladimir1956 [14]

Step 1:

Calculate the measure of angle ∠ABC

\angle DBC+\angle ABC=180(\text{ sum of angles on a straight line)}\angle ABC=65^0\begin{gathered} \angle DBC+\angle ABC=180 \\ \angle DBC+65^0=180^0 \\ \angle DBC=180^0-65^0 \\ \angle DBC=115^0 \end{gathered}

From the triangle in the question,

a=10\operatorname{km},c=15\operatorname{km},B=115^0

Step 2:

Calculate the value of AB using the cosine rule below

b^2=a^2+c^2-2\times a\times c\times\cos B

By substituting the values, we will have

\begin{gathered} b^2=a^2+c^2-2\times a\times c\times\cos B \\ b^2=10^2+15^2-2\times10\times15\times\cos 115^0 \\ b^2=100+225-300\times(-0.4226) \\ b^2=325+126.78 \\ b^2=451.78 \\ \text{Square root both sides} \\ \sqrt[]{b^2}=\sqrt[]{451.78} \\ b=21.26\operatorname{km} \end{gathered}

Hence,

The distance of point A to point C is = 21.26km

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1 year ago
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Here is for solving for x

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