All categories

2 years ago

[(-2)²×3]÷[(-9)-5+2]

Mathematics

2 answers:

Law Incorporation [45]2 years ago

8 0

-1 is the correct answer

Elodia [21]2 years ago

6 0

Answer:

-1

Step-by-step explanation:

1. To simplify this expression, let's go through the steps of BODMAS.

$\frac{(-2)^2(3)}{-9-5+2}$
$\frac{(4)(3)}{-9-5+2}$
$\frac{12}{-9-5+2}$
$\frac{12}{-14+2}$
$\frac{12}{-12}$
$-1$

Therefore, the answer is -1.

You might be interested in

Renee was able to plant 15 plant in 5min at this rate how many flowers could she paint plant in 45 min

Sedbober [7]

Answer:

Answer is 135 plant

Step-by-step explanation:

We have given that,

In 5 min Renee can plant 15 plant.

That means in one minute Renee can plant $\frac{15}{5}$ plant.

i.e. Rate of planting = 3 plant per minute

So, In 45 minutes , she can plant 45×3 plant

i.e Plant she can plant in 45 minute = 135 plant

6 0

3 years ago

What is the domain of the function y=?x-1? 00 0 -1 00 OOO 1

SVETLANKA909090 [29]

Answer:

The set of all real numbers

Step-by-step explanation:

The domain is the set of all real numbers (both positives and negatives including zero.

3 0

3 years ago

Consider the graph of f(x) = x is shifted up 8 units, what would be the equation of the new graph?

anygoal [31]

F(x) = x + 8

This is because you are simply moving the y-intercept up 8 units.

Please consider marking this answer as Brainliest to help me advance.

5 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

For Bus:

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

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

For motorcycle:

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

∴

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

What is [nine x squared multiplied by thirty six y squared plus nine x squared multiplied by 4z squared?

Whitepunk [10]

Answer:

loading, please wait...

Step-by-step explanation:

8 0

3 years ago