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3 years ago

Which best explains why 5 x 4 is greater than 5 x 1/4 ?

1 answer:

3 0

Because one is put into fraction form. Therefore the one in fraction form is going to be smaller. 5*4=20 and 5*1/4=1.25. I hope my answer helped you.

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3 - x3 > -4 help pls

Kamila [148]

Answer:

x<1.912931

Step-by-step explanation:

6 0

3 years ago

A farmer harvested 21 1/4 fewer bushels of sweet corn this year than the previous year. If the farmer's sweet corn crop is plant

klasskru [66]

Answer: 85/12 fewer bushels per acre

Step-by-step explanation:

Hi, to answer this question we have to write an expression with the information given and solve it:

We simply have to divide the total loss (21 1/4) by the number of acres (3):

Mathematically speaking:

21 1/4 ÷3 = (21(4) +1)/4 ÷3 = 85/4 ÷ 3= 85/12 fewer bushels per acre

Feel free to ask for more if needed or if you did not understand something

5 0

3 years ago

Since you have 15% off of $24.99, you should have found (on the previous Slide) that you will have to pay 85% of $24.99. Using t

Ahat [919]

Answer: $21.24

Step-by-step explanation:

Well if you apply the discount you will have to pay 85% of the price which is already said .

So 85% of 24.99 will give you the cost of the shirt.

0.85 * 24.99 = 21.2415

TO the nearest penny will be 21.24

7 0

3 years ago

Suiting at 6 a.m., cars, buses, and motorcycles arrive at a highway loll booth according to independent Poisson processes. Cars

dem82 [27]

Answer:

Step-by-step explanation:

From the information given:

the rate of the cars = $\dfrac{1}{5} \ car / min = 0.2 \ car /min$

the rate of the buses = $\dfrac{1}{10} \ bus / min = 0.1 \ bus /min$

the rate of motorcycle = $\dfrac{1}{30} \ motorcycle / min = 0.0333 \ motorcycle /min$

The probability of any event at a given time t can be expressed as:

$P(event \ (x) \ in \ time \ (t)\ min) = \dfrac{e^{-rate \times t}\times (rate \times t)^x}{x!}$

∴

(a)

$P(2 \ car \ in \ 20 \ min) = \dfrac{e^{-0.20\times 20}\times (0.2 \times 20)^2}{2!}$

$P(2 \ car \ in \ 20 \ min) =0.1465$

$P ( 1 \ motorcycle \ in \ 20 \ min) = \dfrac{e^{-0.0333\times 20}\times (0.0333 \times 20)^1}{1!}$

$P ( 1 \ motorcycle \ in \ 20 \ min) = 0.3422$

$P ( 0 \ buses \ in \ 20 \ min) = \dfrac{e^{-0.1\times 20}\times (0.1 \times 20)^0}{0!}$

$P ( 0 \ buses \ in \ 20 \ min) = 0.1353$

Thus;

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.1465 × 0.3422 × 0.1353

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.0068

(b)

the rate of the total vehicles = 0.2 + 0.1 + 0.0333 = 0.3333

the rate of vehicles with exact change = rate of total vehicles × P(exact change)

$= 0.3333 \times \dfrac{1}{4}$

= 0.0833

∴

$P(zero \ exact \ change \ in \ 10 minutes) = \dfrac{e^{-0.0833\times 10}\times (0.0833 \times 10)^0}{0!}$

P(zero exact change in 10 minutes) = 0.4347

The probability of the 7th motorcycle after the arrival of the third motorcycle is:

$P( 4 \ motorcyles \ in \ 45 \ minutes) =\dfrac{e^{-0.0333\times 45}\times (0.0333 \times 45)^4}{4!}$

$P( 4 \ motorcyles \ in \ 45 \ minutes) =0.0469$

Thus; the probability of the 7th motorcycle after the arrival of the third one is = 0.0469

P(at least one other vehicle arrives between 3rd and 4th car arrival)

= 1 - P(no other vehicle arrives between 3rd and 4th car arrival)

The 3rd car arrives at 15 minutes

The 4th car arrives at 20 minutes

The interval between the two = 5 minutes

<u>For Bus:</u>

P(no other vehicle other vehicle arrives within 5 minutes is)

= $\dfrac{6}{12} = 0.5$

<u>For motorcycle:</u>

$= \dfrac{2 }{12} = \dfrac{1 }{6}$

∴

The required probability = $1 - \Bigg ( \dfrac{e^{-0.5 \times 0.5^0}}{0!} \times \dfrac{e^{-1/6}\times (1/6)^0}{0!} \Bigg)$

= 1- 0.5134

= 0.4866

6 0

3 years ago

Look at the Employee Time Card for

pshichka [43]

I believe it’s 84.50

8 0

3 years ago