All categories

3 years ago

What is the axis of symmetry for the function f(x)=7−4x+x2?

Mathematics

2 answers:

Rufina [12.5K]3 years ago

6 0

Answer:

Answer:

x = 2

Step-by-step explanation:

The vertex form of the parabola is given by

Comparing this equation with vertex form of a parabola , where (h,k) is the vertex

h = 2

k = 3

Hence, vertex is (2,3)

Now, axis of symmetry of a parabola passes through the vertex and divide the parabola in two equal halves.

Hence, axis of symmetry of the parabola is given by

x = h

x = 2

Third option is correct.

Ulleksa [173]3 years ago

3 0

I was going to help you with this one but it looks like you got it all covered !!!!

You might be interested in

Bill measures the circumference of a circular fish pond to be 28 ft.

-Dominant- [34]

9ft? to get the circumference, we multiply the diameter to pi, so all we need to do is divide 28 ÷ π, and round.

5 0

2 years ago

Read 2 more answers

Plz help with this question.If can't do it don't answer.

nekit [7.7K]

Answer:

the total surface area of this triangular prism is 976cm2

6 0

3 years ago

Read 2 more answers

What's -2x-8y=12 in y intercept form

Free_Kalibri [48]

Y=2/-8x-1.5 ( pretty much add 2x to both sides, then divide all parts by negative 8. So it'd be 2 over -8 and then 12 divided by -8 which equals -1.5)

7 0

3 years ago

Please help guys ♡♡♡♡♡♡

kakasveta [241]

Answer:

hope this helped uh

if u like then u can follow me

7 0

3 years ago

Read 2 more answers

A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. Assume that the distr

Marianna [84]

Answer:

(a) Decision rule for 0.01 significance level is that we will reject our null hypothesis if the test statistics does not lie between t = -2.651 and t = 2.651.

(b) The value of t test statistics is 1.890.

(c) We conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) P-value of the test statistics is 0.0662.

Step-by-step explanation:

We are given that a recent study focused on the number of times men and women who live alone buy take-out dinner in a month.

Also, following information is given below;

Statistic : Men Women

The sample mean : 24.51 22.69

Sample standard deviation : 4.48 3.86

Sample size : 35 40

Let $\mu_1$ = mean number of times men order take-out dinners in a month.

$\mu_2$ = mean number of times women order take-out dinners in a month

(a) So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2$ = 0 {means that there is no difference in the mean number of times men and women order take-out dinners in a month}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2\neq$ 0 {means that there is difference in the mean number of times men and women order take-out dinners in a month}

The test statistics that would be used here Two-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_-_n_2_-_2$

where, $\bar X_1$ = sample mean for men = 24.51

$\bar X_2$ = sample mean for women = 22.69

$s_1$ = sample standard deviation for men = 4.48

$s_2$ = sample standard deviation for women = 3.86

$n_1$ = sample of men = 35

$n_2$ = sample of women = 40

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(35-1)\times 4.48^{2}+(40-1)\times 3.86^{2} }{35+40-2} }$ = 4.16

So, test statistics = $\frac{(24.51-22.69)-(0)}{4.16 \sqrt{\frac{1}{35}+\frac{1}{40} } }$ ~ $t_7_3$

= 1.890

(b) The value of t test statistics is 1.890.

(c) Now, at 0.01 significance level the t table gives critical values of -2.651 and 2.651 at 73 degree of freedom for two-tailed test.

Since our test statistics lies within the range of critical values of t, 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 there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) Now, the P-value of the test statistics is given by;

P-value = P( $t_7_3$ > 1.89) = 0.0331

So, P-value for two tailed test is = 2 $\times$ 0.0331 = 0.0662

4 0

3 years ago