The pool has a diameter of 5 feet and a height of 3 feet

A quadratic equation is set up in the form of ax² + bx +c
First set equation = to 0
0= x² - 5x - 24
Next plug into quadratic formula( -b Plus or minus the √b²-4ac) ÷ 2a
[10 plus or minus √(25² - 4×1×24)] ÷ 2
Solve for inner parenthesis first
√625- 96 = √529
Now set up two equations
(10 + √529) ÷2 = x = 16.5
(10 - √529) ÷2 = x = -6.5
So therefore x = 16.5 and -6.5

8 0

3 years ago

What is the area of this figure?

lesantik [10]

Answer:

$\Huge\boxed {A =859ft^{2} }$

Step-by-step explanation:

Hello There!

To solve for the area of this figure we need to split the figure into 3 different parts:

A rectangle with a length of 9 ft and a width of 7 ft

a rectangle with a width of 9ft + 7ft and a length of 25 ft

a rectangle with a width of 18 ft and a length of 22 ft

To find the area of a rectangle we use the formula

$A=w*l$ where w = width and l = length

for the first one we plug in the values

A = 9 * 7

9 * 7 = 63 so the area of the smallest rectangle is 63ft²

Now lets find the area of the larger rectangle

The dimensions are l = 25 ft and w = 9 + 7 (16 ft)

Now we can plug in the values into the area formula

A = 16 * 25

16*25=400 so the area of the larger rectangle is 400 ft²

Now lets find the area of the last rectangle

The dimensions are l = 22 ft and w = 18 ft

now lets plug in the values to the formula

A = 22 * 18 =396

so the area of the last rectangle is 396 ft²

Finally we want to add all of the areas together

396 + 400 + 63 = 859

So the area of the figure is 859 ft²

7 0

3 years ago

Read 2 more answers

A person purchased a slot machine and tested it by playing it 1,137 times. There are 10 different categories of outcomes, includ

ivann1987 [24]

Answer:

$p_v= P(\chi^2_{9}>11.517)=0.2419$

And on this case if we see the significance level given $\alpha=0.1$ we see that $p_v>alpha$ so we fail to reject the null hypothesis that the observed outcomes agree with the expected frequencies at 10% of significance.

Step-by-step explanation:

A chi-square goodness of fit test determines if a sample data obtained fit to a specified population.

$p_v$ represent the p value for the test

O= obserbed values

E= expected values

The system of hypothesis for this case are:

Null hypothesis: $O_i = E_i[/tex[Alternative hypothesis: [tex]O_i \neq E_i$

The statistic to check the hypothesis is given by:

$\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

On this case after calculate the statistic they got: $\chi^2 = 11.517$

And in order to calculate the p value we need to find first the degrees of freedom given by:

$df=n-1=10-1=9$ , where k represent the number of levels (on this cas we have 10 categories)

And in order to calculate the p value we need to calculate the following probability:

$p_v= P(\chi^2_{9}>11.517)=0.2419$

And on this case if we see the significance level given $\alpha=0.1$ we see that $p_v>alpha$ so we fail to reject the null hypothesis that the observed outcomes agree with the expected frequencies at 10% of significance.

6 0

4 years ago