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2 years ago

Subtracting mixed numbers

Mathematics

2 answers:

Marizza181 [45]2 years ago

6 0

Bro this is so easy. Look at some khan academy

liraira [26]2 years ago

3 0

Just listen to ur teacher

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The house numbers on the North side of Flynn Street are even. In one block, the house numbers begin 1022, 1032, 1042, 1052. What

Gnom [1K]

The series 1022,1032,1042,1052....

a1=a=1022
d=a2-a1=1032-1022=10
n=5 as 5th term

Sn= a+(n-1)d
= 1022+(5-1)10
= 1022+40
= 1062
Next house will have 1062 as number.

3 0

3 years ago

3yd 4yd what is the length of the hypotenuse

hjlf

Answer:

5 yd

Step-by-step explanation:

Use Pythagorean theorem,

Hypotenuse² = base² + altitude²

= 3² + 4²

= 9 + 16

= 25

Hypotenuse = √25 = √5*5 = 5 yd

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2 years ago

Which is the most reasonable estimation for the weight of a kitten?

Zielflug [23.3K]

Depending on the age of the kitten they can weigh anywhere between 1 to 2 1/2 pounds or even 3 pounds if its a big kitten. Hope this answers your question!:)

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2 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

T=pd/2x solve for X i always get confused when theirs more then one letter

Ainat [17]

When there's more than one letter, and you just have to solve for one variable, you can just solve it like you would any other equation. Treat the other variables like numbers and add, subtract, multiply, and divide them to both sides of the equation in order to isolate the variable you want to solve for.

$t= \frac{pd}{2x} \\ \\ t*2x= \frac{pd}{2x}*2x \\ \\ 2tx=pd \\ \\ \frac{2tx}{2t} = \frac{pd}{2t} \\ \\ x=\frac{pd}{2t}$

7 0

3 years ago

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