Answer please..<br>help me to solve this..

Help need for test. What is the answer. There is an error in option c. It’s supposed to say line EF not line FF. Will give first

frez [133]

The answer is C I believe

7 0

3 years ago

If AC=5cm, BC=12cm, and m AC= 40 degrees what is the radius of the circumscribed circle

melamori03 [73]

Let's assume they meant C=40 degrees. With an angle like that they're asking for approximation; we'll oblige.

The circumradius is the product of the triangle sides divided by four times the area.

Here we have remaining side given by the Law of Cosines.

$AB^2 = AC^2 + BC^2 - 2\ AC \ BC \cos C$

$AB^2 = AC^2 + BC^2 - 2 AC \ AB \cos C = 5^2 + 12^2 - 2(5)(12) \cos 40^\circ$

$AB = \sqrt{ 169 - 120 \cos 40 ^\circ} \approx 8.77921789$

The area is $\frac 1 2\ AC \ BC \sin C = \frac 1 2 (5)(12) \sin 40^\circ \approx 19.283628$

The circumradius is $r \approx \dfrac{(5)(12)(8.77921789 )}{ 4 (19.283628) } = 6.829019329$

5 0

3 years ago

Vertex form is f(x)=a(x-p)^2 +q. How do i determine a?

slamgirl [31]

Y1 is the simplest parabola. Its vertex is at (0,0) and it passes thru (2,4). This is enough info to conclude that y1 = x^2.

y4, the lower red graph, is a bit more of a challenge. We can easily identify its vertex, which is (-4,0), and several points on the grah, such as (2,-3).

Let's try this: assume that the general equation for a parabola is
y-k = a(x-h)^2, where (h,k) is the vertex. Subst. the known values,

-3-(-4) = a(2-0)^2. Then 1 = a(2)^2, or 1 = 4a, or a = 1/4.

The equation of parabola y4 is y+4 = (1/4)x^2

Or you could elim. the fraction and write the eqn as 4y+16=x^2, or

4y = x^2-16, or y = (1/4)x - 4. Take your pick! Hope this helps you find "a" for the other parabolas.

8 0

3 years ago

A park is mapped on a coordinate plane, where C1, C2, C3, and C4 represent chairs and SW1, SW2, and SW3 represent swings. How fa

serg [7]

Answer:

Option (B)

Step-by-step explanation:

To calculate the distance between C2 and SW1 we will use the formula of distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ .

d = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^2 }$

Coordinates representing positions of C2 and SW1 are (2, 2) and (-6, -7) respectively.

By substituting these coordinates in the formula,

Distance between these points = $\sqrt{(-6-2)^2+(-7-2)^2}$

= $\sqrt{(64)+(81)}$

= $\sqrt{145}$ units

Therefore, Option (B) will be the correct option.

5 0

3 years ago

Read 2 more answers

Many high school students take the AP tests in different subject areas. In 2007, of the 144,796 students who took the biology ex

Nataly [62]

Answer:

$(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942$

$(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978$

And the 90% confidence interval would be given (0.0942;0.09978).

We are confident at 90% that the difference between the two proportions is between $0.0942 \leq p_A -p_B \leq 0.09978$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

$p_A$ represent the real population proportion female for Biology

$\hat p_A =\frac{84199}{144796}=0.582$ represent the estimated proportion female for biology

$n_A=144796$ is the sample size for A

$p_B$ represent the real population proportion female for calculus AB

$\hat p_B =\frac{102598}{211693}=0.485$ represent the estimated proportion female for Calculus AB

$n_B=211693$ is the sample size required for B

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 90% confidence interval the value of $\alpha=1-0.90=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942$

$(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978$

And the 90% confidence interval would be given (0.0942;0.09978).

We are confident at 90% that the difference between the two proportions is between $0.0942 \leq p_A -p_B \leq 0.09978$

5 0

3 years ago

Answer please..help me to solve this..​

Answer please..help me to solve this..