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Aleonysh [2.5K]

2 years ago

9

Points E, F, and D are on circle C, and angle G measures 60°. The measure of arc EF equals the measure of arc FD.

1 answer:

elena-s [515]2 years ago

7 0

∠EFD ≅ ∠EGD

∠EGD ≅ ∠ECDf

Arc E D is-congruent-to arc F D

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P(x)=5x^2-3x+7(x=1)<br> Pls

kramer

Answer:

9

Step-by-step explanation:

Given

p(x) = 5x² - 3x + 7 ← substitute x = 1

p(1) = 5(1)² - 3(1) + 7 = 5(1) - 3(1) + 7 = 5 - 3 + 7 = 9

4 0

3 years ago

An environment engineer measures the amount ( by weight) of particulate pollution in air samples ( of a certain volume ) collect

Serggg [28]

Answer:

$k = 1$

$P(x > 3y) = \frac{2}{3}$

Step-by-step explanation:

Given

$f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x } & { \text 0, { elsewhere. } } \end{array} \right.$

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1$

Where

${ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x }$

Substitute values for the interval of x and y respectively

So, we have:

$\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1$

Isolate k

$k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1$

Integrate y, leave x:

$k \int\limits^2_{0} y {dx} \, [0,x/2]= 1$

Substitute 0 and x/2 for y

$k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1$

$k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1$

Integrate x

$k * \frac{x^2}{2*2} [0,2]= 1$

$k * \frac{x^2}{4} [0,2]= 1$

Substitute 0 and 2 for x

$k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1$

$k *[ \frac{4}{4} - \frac{0}{4} ]= 1$

$k *[ 1-0 ]= 1$

$k *[ 1]= 1$

$k = 1$

Solving (b): $P(x > 3y)$

We have:

$f(x,y) = k$

Where $k = 1$

$f(x,y) = 1$

To find $P(x > 3y)$ , we use:

$\int\limits^a_b \int\limits^a_b {f(x,y)}$

So, we have:

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 dxdy$

Integrate x leave y

$P(x > 3y) = \int\limits^2_0 x [0,y/3]dy$

Substitute 0 and y/3 for x

$P(x > 3y) = \int\limits^2_0 [y/3 - 0]dy$

$P(x > 3y) = \int\limits^2_0 y/3\ dy$

Integrate

$P(x > 3y) = \frac{y^2}{2*3} [0,2]$

$P(x > 3y) = \frac{y^2}{6} [0,2]\\$

Substitute 0 and 2 for y

$P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}$

$P(x > 3y) = \frac{4}{6} -\frac{0}{6}$

$P(x > 3y) = \frac{4}{6}$

$P(x > 3y) = \frac{2}{3}$

8 0

2 years ago

Can someone help me ?

vfiekz [6]

Answer:

The last answer choice is correct, I am pretty sure.

Just go on the calculator (like Desmos) and just use the decimals to determine which one is true.

8 0

3 years ago

Read 2 more answers

Bonds from the city of tavel gorge worth $1,000 each are selling at 110.611. if sean wishes to purchase three such bonds, how mu

umka2103 [35]

The amount that will be expected to be paid for the bonds that are bought will be C. $3,318.33

<h3>How to calculate the amount of bonds?</h3>

From the information given, the bonds from the city of tavel gorge worth $1,000 each are selling at 110.611 and Sean wishes to purchase three such bonds.

Therefore, the amount for the bonds will be:

= 1110.611 × 3

= 3,318.33

In conclusion, the correct option is C.

Learn more about bonds on:

brainly.com/question/2775110

#SPJ4

8 0

1 year ago

20 POINTS ANSWER FAST

Monica [59]

I’m pretty sure the answer is (2,1)

8 0

2 years ago

Read 2 more answers