Shannon

Find the perimeter of a rectangle with a length of

Luba_88 [7]

Answer:

$P = 8\sqrt{2} + 10 + 2\sqrt{6}$

Can only be simplified through calculator using decimal equivalents.

Step-by-step explanation:

The perimeter of a rectangle is the formula P=2l+2w. Here $l=4\sqrt{2}$ and $w=5+\sqrt{6}$ . So the perimeter is:

$P= 2l+2w\\P = 2(4\sqrt{2} ) + 2(5+\sqrt{6} )\\P = 8\sqrt{2} + 10 + 2\sqrt{6}$

4 0

3 years ago

Consider triangle Wes the legs each have a length of 10 units. What is the length of the hypotenuse of the triangle

Orlov [11]

About 14 units hope this helps

7 0

3 years ago

URGENT! HELP PLEASEEE!! ITS DUE SOON

Kazeer [188]

Answer: the answer is D

Step-by-step explanation:

This is because the x and y values constantly add two, making it exponential

8 0

3 years ago

To best estimate the quotient in scientific notation. what number should replace m?

fredd [130]

Answer:

m = 4

Step-by-step explanation:

The complete question is:

Given:

6.33 x 10 ⁹ / 1.79 x 10⁵ ≈ 3 x $10^{m}$

To find:

To best estimate the quotient in scientific notation. what number should replace m?

Solution:

The equation is:

6.33 x 10 ⁹ / 1.79 x 10⁵ = 3 x $10^{m}$

Let x = 9 and y = 5 then the above equation becomes:

6.33 x $10^{x}$ / 1.79 x $10^{y}$ = 3 x $10^{m}$

Since we know that the exponent property is:

quotient $quotient \frac{z^{x} }{z^{y} } = quotient z^{x-y}$

Now converting the given equation in the form above we get:

6.33 / 1.79 x $10^{x-y}$

6.33 / 1.79 x $10^{9-5}$

Now the quotient is 6.33/1.79 ≈ 3.536313

quotient x $10^{9-5}$

3.536313 x $10^{9-5}$

Since 9-5 = 4 So

3.536313 x $10^{4}$

3.536 x $10^{4}$

Hence

6.33 x 10 ⁹ / 1.79 x 10⁵ ≈ 3.5 x $10^{4}$

Hence m = 4

7 0

3 years ago

Suppose that the speed at which cars go on the freeway is normally distributed with mean 68 mph and standard deviation 5 miles p

denis23 [38]

Answer:

A) $X \sim N(68 , 25)$

B) the probability that is traveling more than 70 mph is 0.3446

C) the probability that it is traveling between 65 and 75 mph is 0.6449

D) 90% of all cars travel at least 56.85 mph fast on the freeway

Step-by-step explanation:

The speed at which cars go on the freeway is normally distributed with mean 68 mph and standard deviation 5 miles per hour.

Mean = $\mu = 68 mph$

Standard deviation = $\sigma = 5 mph$

A) X ~ N( _____, _______ )

In general $X \sim N( \mu , \sigma^2)$

$\mu = 68 mph$

$\sigma = 5 mph$

$\sigma^2 = 5^2 = 25$

So, $X \sim N(68 , 25)$

B) If one car is randomly chosen, find the probability that is traveling more than 70 mph.i.e.P(X>70)

So, $Z = \frac{x-\mu}{\sigma}\\Z=\frac{70-68}{5}$

Z=0.4

Using Z table

P(Z>70)=1-P(Z<70)=1-0.6554=0.3446

Hence the probability that is traveling more than 70 mph is 0.3446

C) If one of the cars is randomly chosen, find the probability that it is traveling between 65 and 75 mph.

P(65<X<75)

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

AT x = 65

$Z=\frac{65-68}{5}$

Z=-0.6

AT x = 75

$Z=\frac{75-68}{5}$

Z=1.4

Using Z table

P(65<X<75)=P(-0.6<Z<1.4)=P(Z<1.4)-P(Z<-0.6)=0.9192-0.2743=0.6449

Hence the probability that it is traveling between 65 and 75 mph is 0.6449

D)90% of all cars travel at least how fast on the freeway?

Since we are supposed to find at least how fast on the freeway

So,P(X>x)=0.9

1-P(X<x)=0.9

1-0.9=P(X<x)

0.1=P(X<x)

Z value at 10% =-2.23

So, $Z=\frac{x-\mu}{\sigma}\\-2.23=\frac{x-68}{5}\\-2.23 \times 5 =x-68\\(-2.23 \times 5)+68=x$

56.85 = x

Hence 90% of all cars travel at least 56.85 mph fast on the freeway

8 0

3 years ago