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3 years ago

The length of a rectangle is 2 ft longer than its width.

1 answer:

6 0

If the length is 2ft longer than it’s width
Let’s assume width as X and so the length will be X+2
Perimeter=40ft
Putting the values into a equation, we get,
X+2+X=40
=2X+2=40
=2X=38
=X=19
Therefore
Length =X+2=19+2
=21ft
Width=X=19ft
Area of rectangle =lxb
= 21x19
3992ft
HOPE IT WAS HELPFUL :)

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What Ln(z) is answer????!!

dimaraw [331]

Write <em>z</em> in exponential form:

$z=1-i=\sqrt2 e^{-i\frac\pi4}$

Then taking the logarithm, we get

$\mathrm{Ln}(z)=\ln(\sqrt2) + \ln e^{-i\frac\pi4} = \boxed{\ln(\sqrt2)-\dfrac\pi4i}$

so a is the correct answer.

3 0

3 years ago

Can anyone solve this?

anygoal [31]

Answer:

50.4 km (Answer B)

Step-by-step explanation:

From this information, we are able to create a proportion, 2cm/21km= 4.8cm/x km (x representing the actual distance). After cross-multiplying and simplifying, you get an answer of 50.4km. Hope this helps!

6 0

2 years ago

Read 2 more answers

Draw 2 cards from a standard deck of 52 cards. What is the probability that the first card is a queen and the second card is a k

bixtya [17]

Answer:

4/663

Step-by-step explanation:

There are four queens. The probability that the first card is a queen is 4/52.

There are four kings. There are 51 cards left after the first card has been drawn, so the probability that the second card is a king is 4/51.

Therefore, the probability that the first card is a queen AND the second card is a king is:

P = 4/52 × 4/51

P = 4/663

3 0

3 years ago

Complete the table below.

Nana76 [90]

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2 years ago

Assume that military aircraft use ejection seats designed for men weighing between 141.8 lb and 218 lb. If women's weights are

Mariulka [41]

Answer:

$P(141.8$

And we can find this probability with this difference and using the normal standard table:

$P(-0.639$

Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women

Step-by-step explanation:

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(173.6,49.8)$

Where $\mu=173.6$ and $\sigma=49.8$

We are interested on this probability

$P(141.8$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(141.8$

And we can find this probability with this difference and using the normal standard table:

$P(-0.639$

Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women

8 0

3 years ago