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stira [4]
3 years ago
15

What is the slope of the line y = - 5/4x - 5 ?

Mathematics
1 answer:
Jobisdone [24]3 years ago
7 0

Answer:

-5/4

Step-by-step explanation:

Slope is rise over run which means that the -5 will apply to your y values and the 4 will apply to your x values. This slope would have a point move down 5 units and to the right 4 units to get the location of the next point when going left to right. Because the slope contains a negative, the slope will be angled downward left to right. The negative only applies to the 5.

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Solve for X. 20 points !!!!
GarryVolchara [31]

Answer:

4) x = - 12

5) x = 4

Step-by-step explanation:

4)

Since points T and S are mid points of sides ML and MN of triangle MLN.

Therefore, by mid point theorem,

TS is half of LN

so,

x + 26 =  \frac{1}{2} (x + 40) \\  \\ 2(x + 26) =x + 40 \\  \\ 2x + 52 = x + 40 \\  \\ 2x - x = 40 - 52 \\  \\ x =  - 12

5) In the same way x = 4 will be the correct answer in question number 5.

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3 years ago
. Water in a 10 gallon tank is draining at a rate of 2 gallons per hour. Water in a separate tank is filling at a rate of 4 gall
taurus [48]

Answer:

Step-by-step explanation:

Water in a 10 gallon tank is draining at a rate of 2 gallons per hour.

= 10 - 2x

Water in a separate tank is filling at a rate of 4 gallons per hour.

= 10 + 4x

Equating both Equations together

10 - 2x = 10 + 4x

10 - 10 = 4x - 2x

How long until the tanks have the same amount of water?

Let the time = x

7 0
3 years ago
True or false
LuckyWell [14K]

Answer:

True

Step-by-step explanation:

Outliers still are apart of the data, so it will effect the overall measurement because it is different from the other data collected.

Hope that helps!

3 0
3 years ago
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Five and four hundred ninety one thousand
nadya68 [22]

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6 0
2 years ago
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The​ "fear of negative​ evaluation" (FNE) scores for 11 random female students known to have an eating disorder (1) and 14 rando
SOVA2 [1]

Answer:

a. The 95% confidence interval for the difference in mean is C.I. = -3.6 < μ₂ - μ₁ < 4.96

B. We are 95% confident that the difference in mean FNE scores for bulimic and normal students is inside the confidence interval

c.  The assumptions made are;

The variance of the two distributions are equal

Step-by-step explanation:

The given parameters are;

The mean for the 11 students with an eating disorder, \overline x_1 = 13.82

The standard deviation for the 11 students with an eating disorder, s₁ = 4.92

The mean for the 14 students who do not have an eating disorder, \overline x_2 = 13.14

The standard deviation for the 14 students with an eating disorder, s₂ = 5.29

a. The 95% confidence interval for the difference in mean is given as follows;

\left (\bar{x}_{1}- \bar{x}_{2}  \right )\pm t_{\alpha /2}\sqrt{ \hat \sigma^2 \times\left( \dfrac{1}{n_{1}}+\dfrac{1}{n_{2}} \right)}

The pooled standard deviation, is therefore;

\hat{\sigma} =\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}

Therefore;

\hat{\sigma} =\sqrt{\dfrac{\left ( 11-1 \right )\cdot4.92^{2} +\left ( 14-1 \right )\cdot 5.29^{2}}{11+14-2}} \approx 5.13241

Where at degrees of freedom, df = n₁ + n₂ - 2 = 25 - 2 = 23 the critical-t = 2.07

\left (13.82- 13.14  \right )\pm 2.07 \times\sqrt{5.13241^2 *(\dfrac{1}/{11}+\dfrac{1}{14}\right)}

C.I. = \left (13.82- 13.14  \right )\pm 2.07 \times\sqrt{5.13241^2 \times\left(\dfrac{1}{11}+\dfrac{1}{14}\right)}

Therefore, we get;

C.I. = 0.68\pm 4.28

C.I. = -3.6 < μ₂ - μ₁ < 4.96

b. Therefore, given that the confidence interval extends from positive to negative, therefore, there is a possibility that there is no difference between the mean FNE scores for bulimic and normal students

B. We are 95% confident that the difference in mean FNE scores for bulimic and normal students is inside the confidence interval

c.  The assumptions made are;

The variance of the two distributions are equal

6 0
3 years ago
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