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3 years ago

Is a rectangle a square

Mathematics

2 answers:

morpeh [17]3 years ago

4 0

Answer:

depends.

Step-by-step explanation:

all squares are rectangles but not all retangels are squares.

borishaifa [10]3 years ago

3 0

It is technically a square just stretched out correct me if I’m wrong

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Combine. Express your answer in terms of the smaller unit.<br> 3 L - 700 ml = ___ml

TEA [102]

The answer is 2300 ml.

7 0

3 years ago

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Can someone help me with this problem? Thank you!

vesna_86 [32]

Answer: the answer is 180 $in^{2}$

Step-by-step explanation:

Area of a rectangle multiply length times base. 15 x 12 = 180. Hope this helps!

6 0

2 years ago

Please can you help me I'll mark brainiest please and it's being timed!!

Scrat [10]

Answer:

Step-by-step explanation:

The point ( -2 , 3) lies in the 2nd quadrant

Point are generally in the form (x , y). The point you need to plot is ( -2 , 3) which means the value of x = -2 and the value of y = 3.

I have attached an image you can check it out.

8 0

2 years ago

a truck that can carry no more than 6400 lb is being used to transport refrigerators and upright pianos. Each refrigerator weigh

Tems11 [23]

I'm not going to do the entire question, but I'll do the first part.

For the first part, you have to set up an inequality. Use context clues, such as no more than (the one used in this sentence) to figure out what the symbol should be for the inequality.
In this case, because it is NO MORE THAN 6400, it would be:

6400 $\leq$ 250r + 475p

where r represents the # refrigerators, and p represents the # pianos

Now that you have the equation, make the graph, then for the 3rd part, plug in 12 and 8 to r and p to see if it keeps the inequality true.

Hope this helps! :)

7 0

3 years ago

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Suppose that, after measuring the duration of many telephone calls, a telephone company found their data was well-approximated b

Musya8 [376]

Answer:

a) 7.79%

b) 67.03%

c) Cumulative Distribution Function

$P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b$

Step-by-step explanation:

We are given the following in the question:

$p(x) = 0.1 e^{-0.1x}$

where x is the duration of a call, in minutes.

a) P( calls last between 2 and 3 minutes)

$=\displaystyle\int^3_2 p(x)~ dx\\\\= \displaystyle\int^3_20.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^3_2\\\\=-\Big[e^{-0.3}-e^{-0.2}\Big]\\\\= 0.0779\\=7.79\%$

b) P(calls last 4 minutes or more)

$=\displaystyle\int^{\infty}_4 p(x)~ dx\\\\= \displaystyle\int^{\infty}_40.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^{\infty}_4\\\\=-\Big[e^{\infty}-e^{-0.4}\Big]\\\\=-(0- 0.6703)\\= 0.6703\\=67.03\%$

c) cumulative distribution function

$P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b$

6 0

3 years ago