3x squared equals three in standard form

categorize the following graph is linear increasing, linear decreasing, exponential growth, or exponential decay.

OlgaM077 [116]

Answer:

Exponential Decay is the answer

3 0

3 years ago

There are 15 pieces of fruit in a bowl and 9 of them are oranges. What percentage of the pieces of fruit in the bowl are oranges

Pavlova-9 [17]

Answer:

60%

Step-by-step explanation:

7 0

3 years ago

Which lists all the integer solutions of the inequality |x| < 3?

Mrac [35]

If you would like to know which lists all the integer solutions of the inequality |x| < 3, you can calculate this using the following steps:

|x| < 3
- 3 < x < 3
x = {-2, -1, 0, 1, 2}

The correct result would be C. -2, -1, 0, 1, and 2.

6 0

4 years ago

A nutritionist has developed a diet that she claims will help people lose weight. Twelve people were randomly selected to try th

deff fn [24]

The diet which is developed by the nutritionist is effective at the 0.05 level of significance. Claim is agreed.

<h3>When do we use two-sample t-test?</h3>

The two-sample t-test is used to determine if two population means are equal.

A nutritionist has developed a diet that she claims will help people lose weight. In this,

Twelve people were randomly selected to try the diet.
Their weights were recorded prior to beginning the diet and again after 6 months.

Here are the original weights, in pounds, with the weight after 6 months in parentheses.

Before 192 212 171 215 180 207 165 168 190 184 200 196
After 183 196 174 211 160 191 162 175 190 179 189 195

The mean of the weights before 6 moths is,

$\overline X_1=\dfrac{192+ 212 +171 +215 +180 +207 +165 +168 +190 +184 +200 +196 }{12}\\\overline X_1=190$

The mean of the weights after 6 months is,

$\overline X_2=\dfrac{ 183 +196 +174 +211 +160 +191 +162 +175 +190 +179 +189 +195 }{12}\\\overline X_1=183.75$

Standard deviation of both the data is 16.9 and 14.7.

1. Null and Alternative Hypotheses.

The following null and alternative hypotheses need to be tested:

$\begin{array}{ccl} H_0: \mu_1 & = & \mu_2 \\\\ \\\\ H_a: \mu_1 & > & \mu_2 \end{array}$

This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05 and the degrees of freedom are df = 22. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

$df_{Total} = df_1 + df_2 = 11 + 11 = 22$

Hence, it is found that the critical value for this right-tailed test is

$t_c=1.717$ , for α=0.05 and df=22

The rejection region for this right-tailed test is,

$R = \{t: t > 1.717\}$

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$t = \displaystyle \frac{\bar X_1 - \bar X_2}{\sqrt{ \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}(\frac{1}{n_1}+\frac{1}{n_2}) } }$

$\displaystyle \frac{ 190 - 183.75}{\sqrt{ \frac{(12-1)16.9^2 + (12-1)14.7^2}{ 12+12-2}(\frac{1}{ 12}+\frac{1}{ 12}) } } = 0.967$

(4) Decision about the null hypothesis

Since it is observed that t=0.967≤tc=1.717 it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.1721 and since p=0.1721≥0.05p = 0.1721 it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis H₀ is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1 is greater than μ2, at the α=0.05 significance level.

Confidence Interval

The 95% confidence interval is −7.16<μ<19.66

Thus, the diet which is developed by the nutritionist is effective at the 0.05 level of significance. Claim is agreed.

Learn more about the two-sample t-test here;

brainly.com/question/27198724

#SPJ1

8 0

2 years ago

Find, correct to the nearest degree, the three angles of the triangle with the vertices d(0,1,1), e( 2, 4,3) − , and f(1, 2, 1)

Ksju [112]

Well, here's one way to do it at least...

For reference, let 'a' be the side opposite A (segment BC), 'b' be the side opposite B (segment AC) and 'c' be the side opposite C (segment AB). 

Let P=(4,0) be the projection of B onto the x-axis. 
Let Q=(-3,0) be the projection of C onto the x-axis. 

Look at the angle QAC. It has tangent = 5/4 (do you see why?), so angle A is atan(5/4). 

Likewise, angle PAB has tangent = 6/3 = 2, so angle PAB is atan(2). 

Angle A, then, is 180 - atan(5/4) - atan(2) = 65.225. One down, two to go. 

||b|| = sqrt(41) (use Pythagorian Theorum on triangle AQC) 
||c|| = sqrt(45) (use Pythagorian Theorum on triangle APB) 

Using the Law of Cosines... 
||a||^2 = ||b||^2 + ||c||^2 - 2(||b||)(||c||)cos(A) 
||a||^2 = 41 + 45 - 2(sqrt(41))(sqrt(45))(.4191) 
||a||^2 = 86 - 36 
||a||^2 = 50 
||a|| = sqrt(50) 

Now apply the Law of Sines to find the other two angles. 

||b|| / sin(B) = ||a|| / sin(A) 
sqrt(41) / sin(B) = sqrt(50) / .9080 
(.9080)sqrt(41) / sqrt(50) = sin(B) 
.8222 = sin(B) 
asin(.8222) = B 
55.305 = B 

Two down, one to go... 

||c|| / sin(C) = ||a|| / sin(A) 
sqrt(45) / sin(C) = sqrt(50) / .9080 
(.9080)sqrt(45) / sqrt(50) = sin(C) 
.8614 = sin(C) 
asin(.8614) = C 
59.470 = C 

So your three angles are: 

A = 65.225 
B = 55.305 
C = 59.470

8 0

3 years ago