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2 years ago

For the Relation (3,0), (4,-1), (5,-2), (6,-3) what is the range?

Mathematics

1 answer:

marissa [1.9K]2 years ago

8 0

Step-by-step explanation: The range of a relation is the set of all the y-terms of the relation. So here, the range will be the set of all the y-terms.

The y-terms will be the second number inside the ordered pairs. The range is also listed from least to greatest.

Answer is attached in the image provided.

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Given real numbers a,b > 1.

irina [24]

Answer:

FALSE

Step-by-step explanation:

Recall that a function f(x) is of exponential order c, if there exists a constant M such that and a real r such that

$|f(x)|\leq Me^{cx}\;(x\geq r)$

Now, take a = 2.5 and b = 2

The functions

$f(x)=(2.5)^x\;g(x)=2^x$

are both exponential of order 1, since

$\lim_{x \to \infty}\frac{f(x)}{e^x}=\lim_{x \to \infty}\frac{g(x)}{e^x}=0$

but a>b

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3 years ago

Solve this problem using the substitution method 2y-1/2y=2

stealth61 [152]

Answer:

y=3/2and x =2

Step-by-step explanation:

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3 years ago

An article describes an experiment to determine the effectiveness of mushroom compost in removing petroleum contaminants from so

alexgriva [62]

Solution :

Let $p_1$ and $p_2$ represents the proportions of the seeds which germinate among the seeds planted in the soil containing $3\%$ and $5\%$ mushroom compost by weight respectively.

To test the null hypothesis $H_0: p_1=p_2$ against the alternate hypothesis $H_1:p_1 \neq p_2$ .

Let $\hat p_1, \hat p_2$ denotes the respective sample proportions and the $n_1, n_2$ represents the sample size respectively.

$$\hat p_1 = \frac{74}{155} = 0.477419$

$n_1=155$

$$p_2=\frac{86}{155}=0.554839$

$n_2=155$

The test statistic can be written as :

$$z=\frac{(\hat p_1 - \hat p_2)}{\sqrt{\frac{\hat p_1 \times (1-\hat p_1)}{n_1}} + \frac{\hat p_2 \times (1-\hat p_2)}{n_2}}}$

which under $H_0$ follows the standard normal distribution.

We reject $H_0$ at $0.05$ level of significance, if the P-value or if $|z_{obs}|>Z_{0.025}$

Now, the value of the test statistics = -1.368928

The critical value = $\pm 1.959964$

P-value = $P(|z|> z_{obs})= 2 \times P(z< -1.367928)$

$=2 \times 0.085667$

= 0.171335

Since the p-value > 0.05 and $|z_{obs}| \ngtr z_{critical} = 1.959964$ , so we fail to reject $H_0$ at $0.05$ level of significance.

Hence we conclude that the two population proportion are not significantly different.

Conclusion :

There is not sufficient evidence to conclude that the $\text{proportion}$ of the seeds that $\text{germinate differs}$ with the percent of the $\text{mushroom compost}$ in the soil.