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2 years ago

What is the angle VXU?Plzzz help

Mathematics

1 answer:

maxonik [38]2 years ago

7 0

Answer:

∠ VXU = 105°

Step-by-step explanation:

the chord- chord angle VXU is half the sum of the arcs intercepted by the angle and its vertical angle.

The sum of the arcs in a circle = 360° , so

UV = 360° - (130 + 60 + 90)° = 360° - 280° = 80°

Then

∠ VXU = $\frac{1}{2}$ (80 + 130)° = 0.5 × 210° = 105°

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3 years ago

The mean finish time for a yearly amateur auto race was 186.32 minutes with a standard deviation of 0.305 minute. The winning c

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Complete Question

The mean finish time for a yearly amateur auto race was 186.94 minutes with a standard deviation of 0.305 minute. The winning car, driven by Roger, finished in 185.85 minutes. The previous year's race had a mean finishing time of 110.7 with a standard deviation of 0.115 minute. The winning car that year, driven by Karen, finished in 110.48 minutes.

Find their respective z-scores.

Who had the more convincing victory?

A. Roger had a more convincing victory because of a higher z-score.

B. Karen a more convincing victory because of a higher z-score.

C. Roger had a more convincing victory, because of a lower z-score.

D. Karen a more convincing victory because of a lower z-score.

Answer:

The correct option is D

Step-by-step explanation:

From the question we are told that

The mean of the current year is $\mu_c = 186.32 \ minutes$

The standard deviation is $\sigma_c = 0.305 \ minutes$

The time taken by the winning car for the current year is $x = 185.85 \ minutes$

The mean of the previous year is $\mu_p = 110.7 \ minutes$

The standard deviation the previous year is $\sigma_p = 0.115 \ minutes$

The time taken by the winning car for the previous year is $y = 110.48 \ minutes$

Generally the z-score for current year is mathematically evaluated as

$z_c = \frac{x- \mu_c}{\sigma_c }$

=> $z_c = \frac{ 185.85- 186.32}{0.305 }$

=> $z_c = -1.536$

Generally the z-score for previous year is mathematically evaluated as

$z_p = \frac{y- \mu_p}{\sigma_p }$

=> $z_p = \frac{ 110.48-110.7}{0.115 }$

=> $z_p = -1.913$

From the value obtained we see that $z_p < z_c$ , Hence Karen had a more convincing victory because of the lower z -score.

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2 years ago

I need help.. i really want to go sleep.. thank you so much...

Effectus [21]

Answer:

1) True 2) False

Step-by-step explanation:

1) Given $\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$

To verify that the above equality is true or false:

Now find $\sum\limits_{k=0}^8\frac{1}{k+3}$

Expanding the summation we get

$\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3}$ $\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Now find $\sum\limits_{i=3}^{11}\frac{1}{i}$

Expanding the summation we get

$\sum\limits_{i=3}^{11}\frac{1}{i}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Comparing the two series we get,

$\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$ so the given equality is true.

2) Given $\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\sum\limits_{i=1}^3\frac{3i}{i+5}$

Verify the above equality is true or false

Now find $\sum\limits_{k=0}^4\frac{3k+3}{k+6}$

Expanding the summation we get

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3(0)+3}{0+6}+\frac{3(1)+3}{1+6}+\frac{3(2)+3}{2+6}+\frac{3(3)+4}{3+6}+\frac{3(4)+3}{4+6}$

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}+\frac{12}{8}+\frac{15}{10}$

now find $\sum\limits_{i=1}^3\frac{3i}{i+5}$

Expanding the summation we get

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3(0)}{0+5}+\frac{3(1)}{1+5}+\frac{3(2)}{2+5}+\frac{3(3)}{3+5}$

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}$

Comparing the series we get that the given equality is false.

ie, $\sum\limits_{k=0}^4\frac{3k+3}{k+6}\neq\sum\limits_{i=1}^3\frac{3i}{i+5}$

6 0

3 years ago

What is the angle VXU?Plzzz help ​

What is the angle VXU?Plzzz help