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4 years ago

Using side lengths only, could the triangles be similar?

Mathematics

2 answers:

jok3333 [9.3K]4 years ago

5 0

Answer:

No, StartFraction 0.5 Over 1 EndFraction not-equals StartFraction 1 Over 1.5

Step-by-step explanation:

i just took the quiz

pshichka [43]4 years ago

4 0

Answer:

No, StartFraction 0.5 Over 1 EndFraction not-equals StartFraction 1 Over 1.5 EndFraction not-equals StartFraction 1.5 Over 2 EndFraction.

ANSWER 1

Step-by-step explanation:

took the test..

You might be interested in

an ellipse has a center at the origin , a vertex along the major axis at (13,0), and a focus at (12,0). What is the equation of

-BARSIC- [3]

Answer:

The answer to your question is below

Step-by-step explanation:

Data

Center = (0, 0)

Vertex = (13, 0)

Focus = (12, 0)

Process

From the data we know that it is a horizontal ellipse.

1.- Calculate "a", the distance from the center to the vertex.

a = 13

2.- Calculate "c", the distance from the center to the focus

c = 12

3.- Calculate b

Use the Pythagorean theorem to find it

a² = b² + c²

-Solve for b

b² = a² - c²

-Substitution

b² = 13² - 12²

-Simplification

b² = 169 - 144

b² = 25

b = 5

4.- Find the equation of the ellipse

$\frac{x^{2} }{13^{2}} + \frac{y^{2}}{5^{2}} = 1$ or $\frac{x^{2} }{169} + \frac{y^{2}}{25} = 1$

7 0

4 years ago

Jason drove 110 miles in 2 3/4 hours. What was Jason's speed in miles per hour?

REY [17]

Answer:

It is 40 miles per hour.

Step-by-step explanation:

turning the 2 3/4 to 2.75, you can divide 110 by 2.75 to get 40.

brainliest if i am right? uwu

6 0

3 years ago

If f(x) = 5x + 40, what is f(x) when x = -5? og O 8 O 7 O 15

Juli2301 [7.4K]

Answer:

15.

Step-by-step explanation:

5 * - 5 + 40 = 15

8 0

3 years ago

Read 2 more answers

2(2³-2²) Can you also show how you solved it? Thanks.

valina [46]

The answer is 2[2^2(2-1)]=2*4*1=8

7 0

3 years ago

Read 2 more answers

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.

4 0

3 years ago