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3 years ago

Which quadratic function has a wider graph than y=2x^2?

Mathematics

1 answer:

Trava [24]3 years ago

8 0

Answer:

y = 1/2 x²

Step-by-step explanation:

The coefficient of the first term in a quadratic, in our case here, x², will tell us how the graph stretches. This is akin to the slope within the linear graph. Similar to the slope, the smaller the coefficient value, or value of slope m, the shallower the angle.

When discussing quadratics, the larger the coefficient of our x² term, the steeper, and skinnier the graph. If we want to look for a graph that is wider than y = 2x², then we need to find a graph with a coefficient that is less than 2.

Our only option then is

y = 1/2 x²

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Solve for x in each of the equations or inequalities below, and name the property and/or properties used:

nordsb [41]

Answer

a) 3/4 x = 9

x = $\dfrac{9\times 4}{3}$

x = 12

b) 10+ 3x = 5x

2 x = 10

x = 5

c) a + x = b

x = b - a

d) c x = d

x = $\dfrac{d}{c}$

e) $\dfrac{1}{2} x - g < m$

$\dfrac{1}{2} x< m + g$

x < 2( m + g)

f) q + 5 x = 7 x - r

q + r = 2 x

$x = \dfrac{q + r}{2}$

4 0

3 years ago

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

On a test, 74% of the questions are answered correctly. If 111 questions are correct, how many questions are on the test?

Mashcka [7]

Answer:

150

Step-by-step explanation:

You have to divide the correct answers (111) by the total amount of questions (x) in order to find the 74% (0.74)