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3 years ago

Which choice does NOT describe a parallelogram?

Mathematics

2 answers:

weeeeeb [17]3 years ago

7 0

Answer: B. A quadrilateral that has diagonals that do not bisect each other.

Step-by-step explanation:

A parallelogram is a quadrilateral whose opposite sides are equal or congruent and parallel. Also, the opposite sides in parallelogram are equal or congruent and the sum of two adjacent angles is 180 degrees.The diagonals of parallelogram bisect each other.

Therefore by the properties of parallelogram the choice that does NOT describe a parallelogram is " A quadrilateral that has diagonals that do not bisect each other.".

andreev551 [17]3 years ago

4 0

The diagonals cannot both bisect and NOT bisect so we know it is either A or B.
It is in fact true that the diagonals of a parallelogram bisect each other.
LETTER B is false.

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Suppose you are working in an insurance company as a statistician. Your manager asked you to check police records of car acciden

pochemuha

Answer:

(a) 95% confidence interval for the percentage of all car accidents that involve teenage drivers is [0.177 , 0.243].

(b) We are 95% confident that the percentage of all car accidents that involve teenage drivers will lie between 17.7% and 24.3%.

(c) We conclude that the the percentage of teenagers has not changed since you join the company.

(d) We conclude that the the percentage of teenagers has changed since you join the company.

Step-by-step explanation:

We are given that your manager asked you to check police records of car accidents and out of 576 accidents you selected randomly, teenagers were at the wheel in 120 of them.

(a) Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} }}$ ~ N(0,1)

where, $\hat p$ = sample proportion teenage drivers = $\frac{120}{576}$ = 0.21

n = sample of accidents = 576

p = population percentage of all car accidents

Here for constructing 95% confidence interval we have used One-sample z proportion statistics.

So, 95% confidence interval for the population population, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} }}$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }}$ ]

= [ $0.21-1.96 \times {\sqrt{\frac{0.21(1-0.21)}{576} }}$ , $0.21+1.96 \times {\sqrt{\frac{0.21(1-0.21)}{576} }}$ ]

= [0.177 , 0.243]

Therefore, 95% confidence interval for the percentage of all car accidents that involve teenage drivers is [0.177 , 0.243].

(b) We are 95% confident that the percentage of all car accidents that involve teenage drivers will lie between 17.7% and 24.3%.

(c) We are also provided that before you were hired in the company, the percentage of teenagers who where involved in car accidents was 18%.

The manager wants to see if the percentage of teenagers has changed since you join the company.

Let p = percentage of teenagers who where involved in car accidents

So, Null Hypothesis, $H_0$ : p = 18% {means that the percentage of teenagers has not changed since you join the company}

Alternate Hypothesis, $H_A$ : p $\neq$ 18% {means that the percentage of teenagers has changed since you join the company}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} }}$ ~ N(0,1)

where, $\hat p$ = sample proportion teenage drivers = $\frac{120}{576}$ = 0.21

n = sample of accidents = 576

So, test statistics = $\frac{0.21-0.18}{\sqrt{\frac{0.21(1-0.21)}{576} }}$

= 1.768

The value of the sample test statistics is 1.768.

Now at 0.05 significance level, the z table gives critical value of -1.96 and 1.96 for two-tailed test. Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the the percentage of teenagers has not changed since you join the company.

(d) Now at 0.1 significance level, the z table gives critical value of -1.6449 and 1.6449 for two-tailed test. Since our test statistics does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the the percentage of teenagers has changed since you join the company.

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3 years ago

Create a situation that would require both the functions,f(x) and g(x), and is related to the situations you have already outlin

Solnce55 [7]

Answer:

Step-by-step explanation:

PLUTO ANSWER!

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3 years ago

Caden exercises daily by walking on a treadmill. He sets the machine so that he will walk at a steady rate of 3.6 miles per hour

babunello [35]

Answer:

a) d = 3.6t

b) 10.8 miles

c) 1.2 hours

Step-by-step explanation:

For part a:

Slope of a line: y = mx + b

In this case, the equation would be: d = mt,

where m represents the speed Caden is walking at.

Note: b = 0 in this case because when Caden starts walking on the treadmill, he hasn't really covered any distance yet.

Therefore, plugging in a speed of 3.6 miles per hour, the equation is:

d = 3.6t

For part b:

This is a simple instance of plugging in 3 hours for our t value.

Therefore, d = 3.6 (3)

d = 10.8 miles

For part c:

In this case, we are given that d = 4.32 miles.

Therefore, 4.32 = 3.6t

Dividing both sides by 3.6, we get

t = 1.2 hours

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3 years ago

What is the value of x a. 8.125 b .7.25c.6.125 d.3.25

ahrayia [7]

What's the equation with x? I can't solve it without knowing where to find x....

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3 years ago

Read 2 more answers

The speed at which cars travel on the highway has a normal distribution with a mean of 60 km/h and a standard deviation of 5 km/

oee [108]

The z-score of the speed value gives the measure of dispersion of the from

the mean observed speed.

The probability that the speed of a car is between 63 km/h and 75 km/h is

0.273.

The given parameters are;

The mean of the speed of cars on the highway, $\overline x$ = 60 km/h

The standard deviation of the cars on the highway, σ = 5 km/h

Required:

The probability that the speed of a car is between 63 km/h and 75 km/h

Solution;

The z-score for a speed of 63 km/h is given as follows;

$Z=\dfrac{x-\bar x }{\sigma }$

Which gives;

$Z=\dfrac{63-60 }{5 } = 0.6$

From the z-score table, we have;

P(x < 63) = 0.7257

The z-score for a speed of 75 km/h is given as follows;

$Z=\dfrac{75-60 }{5 } = 3$

Which gives, P(x < 75) = 0.9987

The probability that the speed of a car is between 63 km/h and 75 km/h is therefore;

P(63 < x < 75) = P(x < 75) - P(x < 63) = 0.9987 - 0.7257 = 0.273

The probability that the speed of a car is between 63 km/h and 75 km/h is

0.273.

Learn more here:

brainly.com/question/17489087

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2 years ago