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4 years ago

12

A semicircle and a quarter circle are attached to the sides of a rectangle as shown. What is the area of this figure? Use 3.14 f

or pi. PLEASE HELPPP!!

1 answer:

Serjik [45]4 years ago

5 0

Answer:

need picture but i was in same quiz 109

Step-by-step explanation:

You might be interested in

The table below shows all of the possible outcomes for rolling two six-sided number cubes.

Mice21 [21]

C. 1/4

There is a table with 36 possible outcomes. 9 of them are desired outcomes because they have an even number rolled first and then an odd number rolled second (a few random examples from the table would be 2,1, 2,3, 4,1, 4,5, 6,3 etc.)

All we have to do is make a fraction of the desired outcomes over the total number of possible outcomes. 9/36. That can be simplified into 1/4 and <em>voilà</em>, there is your answer.

Hope this helped ッ

3 0

3 years ago

Read 2 more answers

Evaluate the expression when a= 4 and b= -7. <br> -9a+b

Helen [10]

Answer:

-43

Step-by-step explanation:

So first things first you will multiply -9 by 4 and that gives you -36, Next you will add -7 to -36 which gives you -43.

$-9(4)+(-7)\\-36+(-7) \\= -43$

Hope this helps :D

4 0

3 years ago

Read 2 more answers

Which of the following are solutions to the equation below?

sladkih [1.3K]

Answer:

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=3,\:x=-\frac{1}{2}$

Step-by-step explanation:

considering the equation

$2x^2\:-\:4x\:-\:3\:=\:x$

solving

$2x^2\:-\:4x\:-\:3\:=\:x$

$2x^2-4x-3-x=x-x$

$2x^2-5x-3=0$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=2,\:b=-5,\:c=-3:\quad x_{1,\:2}=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

solving

$x=\frac{-\left(-5\right)+\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

$x=\frac{5+\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}$

$x=\frac{5+\sqrt{49}}{2\cdot \:2}$

$x=\frac{5+7}{4}$

$x=3$

also solving

$x=\frac{-\left(-5\right)-\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

$x=\frac{5-\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}$

$x=\frac{5-\sqrt{49}}{4}$

$x=-\frac{2}{4}$

$x=-\frac{1}{2}$

Therefore,

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=3,\:x=-\frac{1}{2}$

7 0

3 years ago

Yellow Cab Taxi charges a $1.75 flat rate for a ride in the cab. In addition to that, they charge $0.45 per mile. Katie has no m

Katena32 [7]

Answer:

.45x + 1.75 < 12

.45x < 10.25

x < 22.7 miles

8 0

3 years ago

It is known that the life of a particular auto transmission follows a normal distribution with mean 72,000 miles and standard de

scoray [572]

Answer:

a) $P(X$

$P(z$

b) $P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

$P(z>-0.583)=1-P(Z$

c) $P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

d) $z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

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

$X \sim N(72000,12000)$

Where $\mu=72000$ and $\sigma=12000$

We are interested on this probability

$P(X$

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:

$P(X$

And we can find this probability using excel or the normal standard table and we got:

$P(z$

Part b

$P(X>65000)$

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:

$P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>-0.583)=1-P(Z$

Part c

$P(X>100000)$

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:

$P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

Part d

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

5 0

3 years ago