(ax-b)(bx+a)=(bx^2+a)a make x subject of the formula

HELP giving brainlest and more points

cupoosta [38]

Answer:

4,000,000,000 meters

Step-by-step explanation:

4 x 10,000,000,000,000 x 0.0001 = 4,000,000,000 meters

4 0

3 years ago

Read 2 more answers

What is the other solution??

Vladimir79 [104]

Answer:

The other solution is 12.

Step-by-step explanation:

Break the solution into groups: (x^2+2x)+(-12x-24)

Factor out the x^2 from the first group to get x(x+2)

Factor out the -12 out of the second group to get -12(x-12)

Here we get (x-12)(x+2)

You can see solutions are -2 and 12.

4 0

3 years ago

Read 2 more answers

Help asap 80 points!

Montano1993 [528]

Answer:

[-3, 3]

Step-by-step explanation:

The function/ graph is defined from -3 to 3, inclusive of those values

3 0

2 years ago

Read 2 more answers

6 x - 2 [ x + 2 ] > 2 - 3 [ x + 3] could u show me step by step on how to do thiz

Dafna11 [192]

Answer:

x > - 3/7

Step-by-step explanation:

6x - 2x - 4 > 2 - 3 (x + 3)

6x - 2x - 4 > 2 - 3x - 9

4x - 4 > 2 - 3x - 9

4x - 4 > - 7 - 3x

4x - 4 + 3x > - 7

7x > - 7 + 4

7x > - 3

8 0

2 years ago

A study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school

nekit [7.7K]

Answer:

We conclude that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids.

Step-by-step explanation:

We are given that a study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois.

Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids.

Let $p_1$ = proportion of Illinois high school freshmen who have used anabolic steroids.

$p_2$ = proportion of Illinois high school seniors who have used anabolic steroids.

SO, Null Hypothesis, $H_0$ : $p_1=p_2$ {means that there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids}

Alternate Hypothesis, $H_A$ : $p_1\neq p_2$ {means that there is a significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids}

The test statistics that would be used here Two-sample z test for proportions;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of high school freshmen who have used anabolic steroids = $\frac{34}{1679}$ = 0.0203

$\hat p_2$ = sample proportion of high school seniors who have used anabolic steroids = $\frac{24}{1366}$ = 0.0176

$n_1$ = sample of high school freshmen = 1679

$n_2$ = sample of high school seniors = 1366

So, the test statistics = $\frac{(0.0203-0.0176)-(0)}{\sqrt{\frac{0.0203(1-0.0203)}{1679}+\frac{0.0176(1-0.0176)}{1366} } }$

= 0.545

The value of z test statistics is 0.545.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistic 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 there is no significant difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids.

3 0

3 years ago