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3 years ago

Is the following graph a function?

Mathematics

1 answer:

alekssr [168]3 years ago

7 0

Check the picture below.

notice that if we run the "vertical line test", the green vertical line there, touches the graph more than once, and for a function, it has to touch it only once, so no dice.

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PLEASE HURRY

mina [271]

Answer:

X ≥ 3

Step-by-step explanation:

Let the number of grapes you'll buy be X, you need to buy at leaF 3 pounds so:

X ≥ 3

Answered by Gauthmath

5 0

3 years ago

- 4/9 times - 5/8 in fractions.

omeli [17]

Answer:

5/8 x 4/9 = 20/72 as simplified as 5/18 in fraction form. 5/8 x 4/9 = 0.2778 in decimal form.

Step-by-step explanation:

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2 years ago

A researcher is interested in estimating the mean weight of a semi tracker truck to determine the potential load capacity. She t

FinnZ [79.3K]

Answer:

(19229.11 ,20770.89)

Step-by-step explanation:

We are given the following information:

Sample size, n = 17

Sample mean = 20,000 pounds

Sample standard deviation = 1,500 pounds

Confidence level = 95%

Significance level = 5% = 0.05

95% Confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 16 and}~\alpha_{0.05} = \pm 2.119$

$20000 \pm 2.119(\displaystyle\frac{1500}{\sqrt{17}} ) = 20000 \pm 770.89 = (19229.11 ,20770.89)$

4 0

3 years ago

Use the normal distribution and the given sample results to complete the test of the given hypotheses. Assume the results come f

AlladinOne [14]

Answer:

$z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{75}}}=2.43$

Now we can calculate the p value with the following probability:

$p_v =P(z>2.43)=0.0075 \approx 0.008$

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true proportion for this case is higher than 0.5

Step-by-step explanation:

Data given and notation

n=75 represent the random sample taken

$\hat p=0.64$ estimated proportion of interest

$p_o=0.5$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to verify if the true proportion is higher than 0.5:

Null hypothesis: $p =0.5$

Alternative hypothesis: $p > 0.5$

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing the info given we got:

$z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{75}}}=2.43$

Now we can calculate the p value with the following probability:

$p_v =P(z>2.43)=0.0075 \approx 0.008$

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true proportion for this case is higher than 0.5

7 0

3 years ago

Find the following measure for this figure.

miv72 [106K]

l=11
r=5

Now

LSA:-

πrl
π(11)(5)
55πunits ²

4 0

2 years ago