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4 years ago

What times what equals 18

1 answer:

7 0

The question is asking to state the two numbers that the product would be 18, base on my calculation, the possible answer would be 2 * 9, 1 * 18 and 3*6. I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications. Have a nice day

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Solve each equation (Polynomial equations)

lesya [120]

${ {e}^{x} }^{2} = \frac{1}{ {e}^{2} } \times {e}^{3x} \\ \Leftrightarrow { {e}^{x} }^{2} = {e}^{ 3x} \times {e}^{ - 2} \\ \Leftrightarrow { {e}^{x} }^{2} = {e}^{3x - 2} \\ \Leftrightarrow {x}^{2} = 3x - 2 \\ \Leftrightarrow {x}^{2} - 3x + 2 = 0 \\ \Leftrightarrow (x - 1)(x - 2) = 0 \\ x = 1 \: \vee \: x = 2 \\ \\ {32}^{ {x}^{2} - 2x } = \frac{1}{ {4}^{x} } \\ \Leftrightarrow {( {2}^{5} )}^{ {x}^{2} - 2x } = {2}^{ - 2x} \\ \Leftrightarrow {2}^{5 {x}^{2} - 10x } = {2}^{ - 2x} \\ \Leftrightarrow 5 {x}^{2} - 10x = - 2x \\ \Leftrightarrow 5 {x}^{2} - 8x = 0 \\ \Leftrightarrow x(5x - 8) = 0 \\ \Leftrightarrow x = 0 \: \vee \: x = \frac{8}{5} \\ \\ { {e}^{x} }^{3} = {2}^{2x} \\ \Leftrightarrow {x}^{3} = ln( {2}^{2x} ) \\ \Leftrightarrow {x}^{3} - 2x ln(2) = 0 \\ \Leftrightarrow x( {x}^{2} - 2 ln(2) ) = 0 \\ \Leftrightarrow x = 0 \:\vee \: {x}^{2} = 2 ln(2) \\ \Leftrightarrow x = 0 \:\vee \: x = \sqrt{2 ln(2) } \:\vee \: x = - \sqrt{2 ln(2) }$

8 0

4 years ago

Lindsey is tracking the progress of her plant’s growth. Today the plan is 5cm high. The plant grows 1.5 cm per day. What will th

rjkz [21]

31.5 cm . multiply 20 times 5 then added the 1.5 cm

4 0

3 years ago

Read 2 more answers

If you make $11.25/hour, how many hours will you need to work to earn $416.25? Please explain how you figured this out.

Andreas93 [3]

37 hours, you divide 11.25 by 416.25

7 0

3 years ago

Read 2 more answers

Test the claim that the mean GPA of night students is larger than 2 at the .025 significance level. The null and alternative hyp

exis [7]

Answer:

$H_0: \, \mu = 2$ .

$H_1:\, \mu > 2$ .

Test statistics: $z \approx 2.582$ .

Critical value: $z_{1 - 0.025} \approx 1.960$ .

Conclusion: reject the null hypothesis.

Step-by-step explanation:

The claim is that the mean $\mu$ is greater than $2$ . This claim should be reflected in the alternative hypothesis:

$H_1:\, \mu > 2$ .

The corresponding null hypothesis would be:

$H_0:\, \mu = 2$ .

In this setup, the null hypothesis $H_0:\, \mu = 2$ suggests that $\mu_0 = 2$ should be the true population mean of GPA.

However, the alternative hypothesis $H_1:\, \mu > 2$ does not agree; this hypothesis suggests that the real population mean should be greater than $\mu_0= 2$ .

One way to test this pair of hypotheses is to sample the population. Assume that the population mean is indeed $\mu_0 = 2$ (i.e., the null hypothesis is true.) How likely would the sample (sample mean $\overline{X} = 2.02$ with sample standard deviation $s = 0.06$ ) be observed in this hypothetical population?

Let $\sigma$ denote the population standard deviation.

Given the large sample size $n = 60$ , the population standard deviation should be approximately equal to that of the sample:

$\sigma \approx s = 0.06$ .

Also because of the large sample size, the central limit theorem implies that $Z= \displaystyle \frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}}$ should be close to a standard normal random variable. Use a $Z$ -test.

Given the observation of $\overline{X} = 2.02$ with sample standard deviation $s = 0.06$ :

$\begin{aligned}z_\text{observed}&= \frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}} \\ &\approx \frac{\overline{X} - \mu_0}{s / \sqrt{n}} = \frac{2.02 - 2}{0.06 / \sqrt{60}} \approx 2.582\end{aligned}$ .

Because the alternative hypothesis suggests that the population mean is greater than $\mu_0 = 2$ , the null hypothesis should be rejected only if the sample mean is too big- not too small. Apply a one-sided right-tailed $z$ -test. The question requested a significant level of $0.025$ . Therefore, the critical value $z_{1 - 0.025}$ should ensure that $P( Z > z_{1 - 0.025}) = 0.025$ .

Look up an inverse $Z$ table. The $z_{1 - 0.025}$ that meets this requirement is $z_{1 - 0.025} \approx 1.960$ .

The $z$ -value observed from the sample is $z_\text{observed}\approx 2.582$ , which is greater than the critical value. In other words, the deviation of the sample from the mean in the null hypothesis is sufficient large, such that the null hypothesis needs to be rejected at this $0.025$ confidence level in favor of the alternative hypothesis.

3 0

3 years ago

The miles-per-gallon obtained by the 1995 model Z cars is normally distributed with a mean of 22 miles-per-gallon and a standard

AlekseyPX

Answer:

0.42074

Step-by-step explanation:

In this question, we are asked to calculate the probability that a car will get less than 21 miles per gallon.

This is P(x < 21)

z = 21-22/5 = -1/5 = -0.2

Therefore the probability of getting less than 21 miles per gallon will be;

P(x<21) = P (z < -0.2)

= 0.42074

4 0

3 years ago