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Svetradugi [14.3K]
3 years ago
15

Which of the following statements about the function cannot be used to conclude that is defined at x = 1?

Mathematics
1 answer:
Murljashka [212]3 years ago
3 0
A Lin x —>1 f(x) exists
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Brody shops at three different grocery stores. He uses the ads to determine where to buy certain items. The table shows the cost
Y_Kistochka [10]

Answer:

The Corner Store sells pb at $0.16 per ounce

Minnie's Market sells pb at $0.14 per ounce

Food Now! sell pb at $0.15 per ounce

And the best place to buy would be Minnie's Market.

Step-by-step explanation:

TCS: $1.28/8=$0.16

MM: $1.96/14=$0.14

FN: $3.30/22=$0.15

7 0
3 years ago
Complete the statement.<br> -6/7 -3/7<br> &lt; &gt;
aalyn [17]
Answer: -9/7



Explanation: subtract both numbers because they are negativeand you get -9/7
4 0
3 years ago
A sample of 100 workers located in Atlanta has an average daily work time of 6.5 hours with a standard deviation of 0.5 hours. A
Readme [11.4K]

Answer:

t=\frac{6.5-6.7}{\sqrt{\frac{0.5^2}{100}+\frac{0.7^2}{110}}}}=-2.398  

df = n_1 +n_2 -2= 100+110-2= 208

Since is a bilateral test the p value would be:

p_v =2*P(t_{208}

Comparing the p value with the significance level given \alpha=0.05 we see that p_v so we can conclude that we can reject the null hypothesis, and we have significant differences between the two groups at 5% of significance.

Step-by-step explanation:

Data given and notation

\bar X_{1}=6.5 represent the sample mean for Atlanta

\bar X_{2}=6.7 represent the sample mean for Chicago

s_{1}=0.5 represent the sample deviation for Atlanta

s_{2}=0.7 represent the sample standard deviation for Chicago

n_{1}=100 sample size for the group Atlanta

n_{2}=110 sample size for the group Chicago

t would represent the statistic (variable of interest)

\alpha=0.01 significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the meanfor atlanta is different from the mean of Chicago, the system of hypothesis would be:

Null hypothesis:\mu_{1}=\mu_{2}

Alternative hypothesis:\mu_{1} \neq \mu_{2}

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

t=\frac{6.5-6.7}{\sqrt{\frac{0.5^2}{100}+\frac{0.7^2}{110}}}}=-2.398  

What is the p-value for this hypothesis test?

The degrees of freedom are given by:

df = n_1 +n_2 -2= 100+110-2= 208

Since is a bilateral test the p value would be:

p_v =2*P(t_{208}

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given \alpha=0.05 we see that p_v so we can conclude that we can reject the null hypothesis, and we have significant differences between the two groups at 5% of significance.

3 0
3 years ago
Read 2 more answers
Can someone please help me out here?
fenix001 [56]
1. D
2. A
3. C
Question 2 MIGHT be D, but I think it's A
6 0
3 years ago
Read 2 more answers
Compute 41, 42, 43, 44, 45, 46, 47, and 48. Make a conjecture about the units digit of 4n where n is a positive integer. Use str
padilas [110]

Answer:

The conjecture is that the unit digit of 4^n = 4 when n = odd also 4^n = 6 when n = even

Step-by-step explanation:

4^1 = 4\\4^2 = 16\\4^3 = 64\\4^4 = 256 \\4^5 = 1024\\4^6 = 4096\\4^7 = 16384\\4^8 = 65536

The conjecture is that the unit digit of 4^n = 4 when n = odd also 4^n = 6 when n = even

To prove this conjecture

4^1 = 4\\4^2 = 16 unit digit = 6

hence the property is true for ; n = 1 and n = 2 and also for every odd and even number ( i.e. from 1  to 8 )

8 0
3 years ago
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