Use the drop-down menus to complete the statements about the function p(x) = x(x

PLS HELP ME ON THIS QUESTION I WILL MARK YOU AS BRAINLIEST IF YOU KNOW THE ANSWER!!

KiRa [710]

Answer:

D :6

Step-by-step explanation:10 is q3 4 is q1subtract and get 6

6 0

3 years ago

F= 8/5 (c) + 40 when c=5

Molodets [167]

Answer:

F = 48

Step-by-step explanation:

F= 8/5 (c) + 40

Let c=5

F = 8/5 * 5 +40

= 8 + 40

= 48

6 0

3 years ago

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Assume that you have a sample of n 1 equals 6, with the sample mean Upper X overbar 1 equals 50, and a sample standard deviati

tigry1 [53]

Answer:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

$df=6+5-2=9$

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

$n_1 =6$ represent the sample size for group 1

$n_2 =5$ represent the sample size for group 2

$\bar X_1 =50$ represent the sample mean for the group 1

$\bar X_2 =38$ represent the sample mean for the group 2

$s_1=7$ represent the sample standard deviation for group 1

$s_2=8$ represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

We are assuming that the population variances for each group are the same

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

The statistic for this case is given by:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

The pooled variance is:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

We can find the pooled variance:

$S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67$

And the pooled deviation is:

$S_p=7.46$

The statistic is given by:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

The degrees of freedom are given by:

$df=6+5-2=9$

The p value is given by:

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

4 0

3 years ago

Pls do it for me thank you so much

Bond [772]

Answer:

(a) Attached to this response.

(b)

(i) 0 cups = 0.275

(ii) 4 cups = 0.125

(iii) 8 cups = 0.025

Step-by-step explanation:

(a) The frequency and relative frequency table has been attached to this response.

i. The first column, labelled x, is the number of cups of coffee consumed per day.

ii. The second column, labelled f, is the number of people that consumed x cups of coffee per day. It is found by counting the number of occurrences (i.e frequency) of the numbers of the first column, x, in the given data.

For example, 0 in first column appears 11 times in the given data. Also, 5 in the first column appears 4 times in the given data.

iii. The third column, labelled r, is the relative relative frequency of the number of cups of coffee. It is calculated by dividing the frequency of the cups of coffee by the total number of people that consumed it. As shown on the table, it is calculated by dividing the each of the values on the second column by the sum of the values on that same column (second column).

For example, the relative frequency of 2 cups of coffee is given by:

r = 7 ÷ 40 = 0.175

Also, the relative frequency of 4 cups of coffee is given by;

r = 5 ÷ 40 = 0.125

(b) The probability (Pₓ) that a randomly selected person consumed x cups of coffee is given by;

Pₓ = frequency of x ÷ total frequency

This is also the relative frequency of x

Therefore,

(i) The probability (P₀) that a randomly selected person consumed 0 cups of coffee is given by;

the relative frequency of 0 = 0.275

(ii) The probability (P₄) that a randomly selected person consumed 4 cups of coffee is given by;

the relative frequency of 4 = 0.125

(iii) The probability (P₈) that a randomly selected person consumed 8 cups of coffee is given by;

the relative frequency of 8 = 0.025

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5 0

3 years ago

Simplify the rational expression. State any restrictions on the variable n^4-11n^2+30/ n^4-7n^2+10

djyliett [7]

To factor both numerator and denominator in this rational expression we are going to substitute $n^{2}$ with $x$ ; so $n^{2} =x$ and $n ^{4} = x^{2}$ . This way we can rewrite the expression as follows:
$\frac{n^{4}-11n^{2} +30 }{n^{2}-7n^{2} +10 } = \frac{ x^{2} -11x+30}{ x^{2} -7x+10}$
Now we have two much easier to factor expressions of the form $a x^{2} +bx+c$ . For the numerator we need to find two numbers whose product is $c$ (30) and its sum $b$ (-11); those numbers are -5 and -6. $(-5)(-6)=30$ and $-5-6=-11$ .
Similarly, for the denominator those numbers are -2 and -5. $(-2)(-5)=10$ and $-2-5=-7$ . Now we can factor both numerator and denominator:
$\frac{ x^{2} -11x+30}{ x^{2} -7x+10} = \frac{(x-6)(x-5)}{(x-2)(x-5)}$
Notice that we have $(x-5)$ in both numerator and denominator, so we can cancel those out:
$\frac{x-6}{x-2}$
But remember than $x= n^{2}$ , so lets replace that to get back to our original variable:
$\frac{n^{2}-6 }{n^{2}-2 }$
Last but not least, the denominator of rational expression can't be zero, so the only restriction in the variable is $n^{2} -2 \neq 0$
$n^{2} \neq 2$
$n \neq +or- \sqrt{2}$

5 0

3 years ago

Read 2 more answers

Use the drop-down menus to complete the statements about the function p(x) = x(x - 1) + 1.