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2 years ago

Evaluate the expression 4−(1−4)−2 =

Mathematics

2 answers:

Dominik [7]2 years ago

6 0

You do 4-(-3)-2 then you get

4+3-2 then u get

Cloud [144]2 years ago

3 0

Parentheses first. 1-4= -3. 4-3=1. 1-2=-1.

The answer is -1.

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What is the interquartile range of this data?

Alexxx [7]

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2 years ago

Read 2 more answers

A person takes a trip, driving with a constant peed of 89.5 km/h, except for a 22.0-min rest stop. If the peron's average speed

Stolb23 [73]

Answer:

a) The person traveled 2.83 hours.

b) The person travels 220.17 kilometers.

Step-by-step explanation:

We have that the speed is the distance divided by the time. Mathematically, that is

$s = \frac{d}{t}$

(a) how much time is spent on the trip and

The peron's average speed is 77.8 km/h, which means that $s = 77.8$

The person distance traveled is:

22 min is 22/60 = 0.37h.

So for the time t1, the person traveled at a speed of 89.5 km/h. Which has a distance of 89.5*t1.

For 0.37h, the person was at a stop, so she did not travel. This means that the total distance is

$d = 89.5t1 + 0 = 89.5t1$

The total time is the time traveling t and the stoppage time 0.37. So

$t = t1 + 0.37$

We want to find t1, which is the time that the person was driving.

$s = \frac{d}{t}$

$77.8 = \frac{89.5t1}{t1 + 0.37}$

$77.8t1 + 77.8*0.37 = 89.5t1$

$11.7t1 = 28.786$

$t1 = \frac{28.786}{11.7}$

$t1 = 2.46$

The total time is

$t = t1 + 0.37 = 2.46 + 0.37 = 2.83$

The person traveled for 2.83 hours.

(b) how far does the person travel?

The person traveled 2.46 hours at an average speed of 77.8 km/h. So

$s = \frac{d}{t}$

$77.8 = \frac{d}{2.83}$

$d = 77.8*2.83 = 220.17$

The person travels 220.17 kilometers.

8 0

2 years ago

What equation is equal to<br> 3/8 less than a number is 11

PtichkaEL [24]

Answer:

$11=x-\frac{3}{8}$

5 0

2 years ago

How do I do this? it doesn't make sense.

yuradex [85]

I think you subtitute de 34 by x not sure

4 0

2 years ago

The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 5 m

bezimeni [28]

Answer: 0.75

Step-by-step explanation:

Given : Interval for uniform distribution : [0 minute, 5 minutes]

The probability density function will be :-

$f(x)=\dfrac{1}{5-0}=\dfrac{1}{5}=0.2\ \ ,\ 0$

The probability that a given class period runs between 50.75 and 51.25 minutes is given by :-

$P(x>1.25)=\int^{5}_{1.25}f(x)\ dx\\\\=(0.2)[x]^{5}_{1.25}\\\\=(0.2)(5-1.25)=0.75$

Hence, the probability that a randomly selected passenger has a waiting time greater than 1.25 minutes = 0.75

7 0

2 years ago