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Brut [27]
3 years ago
10

I need to find the slope and the y intercept of y=-1/4+3 and how to graph it

Mathematics
2 answers:
Vinil7 [7]3 years ago
7 0

Answer:

The y-intercept is:  3   You will start at (0,3)

The slope is: 1/4 meaning that you will go up 1 and 4 to the right each time

Step-by-step explanation:

djverab [1.8K]3 years ago
5 0
The y intercept is 3.
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ANswer this question!!!!!
4vir4ik [10]

<u><em>Its Going to Be Table B! Good Luck :)⇔</em></u>


6 0
3 years ago
IM PUTTING THIS FOR 30 POINTS PLEASE ANSWER
salantis [7]

Answer:

2(Jill) - 20 + (Jill) = 205

3(Jill) = 225

(Jill) = 75

(Jack) = 2(75) - 20 => 150 - 20 = 130

Jill= 75

Jack= 130

Step-by-step explanation:

6 0
3 years ago
Solve for C.<br> 5/4= -4c + 1/4
dybincka [34]
5/4= -4c + 1/4

Inverse operation

\frac{5}{4} = -4c + \frac{1}{4}

- 1/4             -1/4

5/4 - 1/4 = 4/4

\frac{4}{4} = -4c
-4c/-4  = 4/4 ÷ -4/1 

4/4 *- 1/4 = c
-4/8 = c
c = -1/2

Answer:  c = - \frac{1}{2}

6 0
3 years ago
The question is in the picture. will give brainilist.​
sergeinik [125]
Value of y = 30 and value of x = 30.3
3 0
3 years ago
Read 2 more answers
It is desired to compare the hourly rate of an entry-level job in two fast-food chains. Eight locations for each chain are rando
notka56 [123]

Answer:

It can be concluded that at 5% significance level that there is no difference in the amount paid by chain A and chain B for the job under consideration

Step by Step Solution:

The given data are;

Chain A 4.25, 4.75, 3.80, 4.50, 3.90, 5.00, 4.00, 3.80

Chain B 4.60, 4.65, 3.85, 4.00, 4.80, 4.00, 4.50, 3.65

Using the functions of Microsoft Excel, we get;

The mean hourly rate for fast-food Chain A, \overline x_1 = 4.25

The standard deviation hourly rate for fast-food Chain A, s₁ = 0.457478

The mean hourly rate for fast-food Chain B, \overline x_2 = 4.25625

The standard deviation hourly rate for fast-food Chain B, s₂ = 0.429649

The significance level, α = 5%

The null hypothesis, H₀:  \overline x_1 = \overline x_2

The alternative hypothesis, Hₐ:  \overline x_1 ≠ \overline x_2

The pooled variance, S_p^2, is given as follows;

S_p^2 = \dfrac{s_1^2 \cdot (n_1 - 1) + s_2^2\cdot (n_2-1)}{(n_1 - 1)+ (n_2 -1)}

Therefore, we have;

S_p^2 = \dfrac{0.457478^2 \cdot (8 - 1) + 0.429649^2\cdot (8-1)}{(8 - 1)+ (8 -1)} \approx 0.19682

The test statistic is given as follows;

t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{S_{p}^{2} \cdot \left(\dfrac{1 }{n_{1}}+\dfrac{1}{n_{2}}\right)}}

Therefore, we have;

t=\dfrac{(4.25-4.25625)}{\sqrt{0.19682 \times \left(\dfrac{1 }{8}+\dfrac{1}{8}\right)}} \approx -0.028176

The degrees of freedom, df = n₁ + n₂ - 2 = 8 + 8 - 2 = 14

At 5% significance level, the critical t = 2.145

Therefore, given that the absolute value of the test statistic is less than the critical 't', we fail to reject the null hypothesis and it can be concluded that at 5% significance level that chain A pays the same as chain B for the job under consideration

3 0
2 years ago
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