All categories

3 years ago

Which comparison is correct? 140.07 < 140.7 140.07=140.7 140.07 > 140.7

Mathematics

2 answers:

creativ13 [48]3 years ago

7 0

The answer is 140.07< 140.7 I hope I help

Elis [28]3 years ago

6 0

Answer:

40.07 < 140.7

step by step explanation:it's just correct I think

You might be interested in

Simplify

OlgaM077 [116]

Answer:

see below

Step-by-step explanation:

$\frac{x-1}{x^2-3x+2}+ \frac{x-2}{x^2-5x+6} +\frac{x-5}{x^2-8x+15}$

we need to simplify that

$x^2-3x+2=(x-1)(x-2)\\\\x^2-5x+6=(x-2)(x-3)\\\\x^2-8x+15=(x-3)(x-5)$

so we can continue

$\frac{x-1}{(x-1)(x-2)}=\frac{1}{x-2}\\\\\frac{x-2}{(x-2)(x-3)} =\frac{1}{x-3}\\\\\frac{x-5}{(x-3)(x-5)} =\frac{1}{x-3}$

and we can put all together

$\frac{1}{x-2}+ \frac{1}{x-3}+ \frac{1}{x-3}\\\\\frac{1}{x-2} +\frac{2}{x-3}\\\\\frac{x-3}{(x-3)(x-2)}+ \frac{2(x-2)}{(x-2)(x-3)} \\\\\frac{x-3+2x-4}{(x-3)(x-2)}\\\\\frac{3x-7}{x^2-5x+6}$

3 0

3 years ago

A football gained 11 yards on each of 8 plays. How many yards did they gain?

gladu [14]

Answer:he gained 88 yards

Step-by-step explanation: just do 11x 8 and you will get 88

so therefore the best option is a have a great day

4 0

2 years ago

Read 2 more answers

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,

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