Jonathan’s class has 30 boys. Of the students in his class, 60% are girls. How many girls are in Jonathan’s class?

AysviL [449]

(6 • 2 - 2x + 16) + (4 • 2 + 7x - 55)
(-2x + 16 + 6 • 2) + (7x - 55 + 4 • 2)
(-2x + 16 + 12) + (7x - 55 + 8)
-2x + 28 + 7x - 47
-2x + 7x + 28 - 47
5x + (-19)
5x - 19
So the answer is 5x - 19.

8 0

3 years ago

Draw a line representing the "rise" and a line representing the "run" of the line. State the slope of the line in simplest form.

Paul [167]

The slope is 3/4
you just count how many it goes up before it goes over to a whole unit

8 0

3 years ago

17.

777dan777 [17]

Answer:

$12.5

Step-by-step explanation:

Since prize is split equally.

So, money received by each person

= 75/6

= $12.5

7 0

3 years ago

What's the slope of the line passing through the points -1, 8 - 7, 3

tatuchka [14]

Answer:

slope = 5/6

Step-by-step explanation:

Since you were given two points, you can use the point-slope formula to find the slope. The general equation looks like this:

y₁ - y₂ = m(x₁ - x₂)

In this formula, "m" represents the slope. To find the slope, plug the values from the two points into the equation. Make sure to put the values from the same point in the variable with the same number.

Point 1: (-1, 8)

Point 2: (-7, 3)

y₁ - y₂ = m(x₁ - x₂) <----- Original formula

8 - y₂ = m(-1 - x₂) <----- Plug in "x" and "y" values from Point 1

8 - 3 = m(-1 - (-7)) <----- Plug in "x" and "y" values from Point 2

5 = m(-1 - (-7)) <----- Simplify left side

5 = m(6) <----- Simplify inside parentheses

5/6 = m <----- Divide both sides by 6

5 0

2 years ago

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