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VikaD [51]
3 years ago
11

⦁ How can a table be used to find the rate of change and the initial value? Describe the process.

Mathematics
1 answer:
BARSIC [14]3 years ago
8 0

1) Table can be used to find the rate of change and the initial value5

x ------> y

0 -------> 3

1 ------> 5

2 ------> 7

Rate of change (Slope) = \frac{Change in y}{Change in x}

=\frac{ 7 - 5}{2 - 1}

= 2/1 = 2

Rate of change = 2

Y intercept is the point where x=0

So y = 3 from the table.

2) Attached the graph of a line

Rate of change = \frac{Change in y(Rise)}{Change in x} (run)

= \frac{2}{1} = 2

3) slope = \frac{2}{3} and a y-intercept = 7

Equation of a line y= mx + b

Where m is the slope and b is the y intercept

Replace slope m and y intercept

m= 2/3, b= 7

Equation of a line : y = \frac{2}{3} x + 7

4) Carson has $450 in his bank account and deposits $70 per month out of his babysitting money.

Initial amount in the bank (y intercept) b= $450

Deposited amount = 70 that is the slope m= 70

So the equation y = mx+ b , where x is the month

y = 70x + 450

Carson’s bank balance f(x) = 70x + 450

5) the graph of straight line is linear. In the graph x<1 that is (-infinity , 1) is linear.

The graph with curved line is non linear.In the graph x>1 that is (1, infinity ,) is non linear.

The graph is attached below.

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Answer:

14\frac{1}{3}

Step-by-step explanation:

Let's write this out as an equation. Let the unknown quantity be x:

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5) In a certain supermarket, a sample of 60 customers who used a self-service checkout lane averaged 5.2 minutes of checkout tim
Ne4ueva [31]

Answer:

\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643

S_p=2.940

t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751

df=60+72-2=130

p_v =P(t_{130}

Assuming a significance level of \alpha=0.05 we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane

Step-by-step explanation:

Data given

Our notation on this case :

n_1 =60 represent the sample size for people who used a self service

n_2 =72 represent the sample size for people who used a cashier

\bar X_1 =5.2 represent the sample mean for people who used a self service

\bar X_2 =6.1 represent the sample mean people who used a cashier

s_1=3.1 represent the sample standard deviation for people who used a self service

s_2=2.8 represent the sample standard deviation for people who used a cashier

Assumptions

When we have two independent samples from two normal distributions with equal variances we are assuming that  

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

The statistic 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}}}

And t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

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

System of hypothesis

Null hypothesis: \mu_1 \geq \mu_2

Alternative hypothesis: \mu_1 < \mu_2

This system is equivalent to:

Null hypothesis: \mu_1 - \mu_2 \geq 0

Alternative hypothesis: \mu_1 -\mu_2 < 0

We can find the pooled variance:

\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643

And the deviation would be just the square root of the variance:

S_p=2.940

The statistic is given by:

t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751

The degrees of freedom are given by:

df=60+72-2=130

And now we can calculate the p value with:

p_v =P(t_{130}

Assuming a significance level of \alpha=0.05 we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane

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I hope this helps!

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The correct choices are: A, B, C

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