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3 years ago

Simplify 2a^2b^3(4a^2+3ab^2-ab)

1 answer:

3 0

<span>Simplifying 2a2b3(4a2 + 3ab2 + -1ab) = 0 Reorder the terms: 2a2b3(-1ab + 3ab2 + 4a2) = 0 (-1ab * 2a2b3 + 3ab2 * 2a2b3 + 4a2 * 2a2b3) = 0 (-2a3b4 + 6a3b5 + 8a4b3) = 0 Solving -2a3b4 + 6a3b5 + 8a4b3 = 0 Solving for variable 'a'. Factor out the Greatest Common Factor (GCF), '2a3b3'. 2a3b3(-1b + 3b2 + 4a) = 0 Ignore the factor 2. so remember to mark me the brainiest..please</span>

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Find the mean of 4,8,8,9,11

Len [333]

The mean of the given numbers is 8.

7 0

4 years ago

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On Saturday morning, Owen earned $26. By the end of the afternoon he had earned a total of $57. Enter an equation, using x as yo

AnnyKZ [126]

Answer:

$26 + X = $57

$26 + $31 = $57

X = $31

My first time ever answering a question hope it helped =) (And sry it took long)

Step-by-step explanation:

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3 years ago

A random sample of n measurements was selected from a population with unknown mean mu and standard deviation sigmaequals50 for e

Andre45 [30]

Answer:

a) (26.50;57.50)

b) (117.34;128.66)

c) (12.13;27.87)

d) (-4.73;11.01)

e) No. Since the sample sizes are large (n ≥ 30), the central limit theorem guarantees that $\bar x$ is approximately normal, so the confidence intervals are valid

Step-by-step explanation:

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The confidence interval is given by this formula:

$\bar X \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$ (1)

And for a 95% of confidence the significance is given by $\alpha=1-0.95=0.05$ , and $\frac{\alpha}{2}=0.025$ . Since we know the population standard deviation we can calculate the critical value $z_{0.025}= \pm 1.96$

Part a

$n=40,\bar X=42,\sigma=50$

If we use the formula (1) and we replace the values we got:

$42 - 1.96 \frac{50}{\sqrt{40}}=26.50$

$42 + 1.96 \frac{50}{\sqrt{40}}=57.50$

The 95% confidence interval is given by (26.50;57.50)

Part b

$n=300,\bar X=123,\sigma=50$

If we use the formula (1) and we replace the values we got:

$123 - 1.96 \frac{50}{\sqrt{300}}=117.34$

$123 + 1.96 \frac{50}{\sqrt{300}}=128.66$

The 95% confidence interval is given by (117.34;128.66)

Part c

$n=155,\bar X=20,\sigma=50$

If we use the formula (1) and we replace the values we got:

$20 - 1.96 \frac{50}{\sqrt{155}}=12.13$

$20 + 1.96 \frac{50}{\sqrt{155}}=27.87$

The 95% confidence interval is given by (12.13;27.87)

Part d

$n=155,\bar X=3.14,\sigma=50$

If we use the formula (1) and we replace the values we got:

$3.14 - 1.96 \frac{50}{\sqrt{155}}=-4.73$

$3.14 + 1.96 \frac{50}{\sqrt{155}}=11.01$

The 95% confidence interval is given by (-4.73;11.01)

Part e

No. Since the sample sizes are large (n ≥ 30), the central limit theorem guarantees that $\bar x$ is approximately normal, so the confidence intervals are valid

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3 years ago

10.9 as a fraction or mixed number

strojnjashka [21]

The answer is 10 9/10 hope this helped

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3 years ago

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A carton of juice contains. 64 ounces. Ms Wilson brought 6. of juice how many ounces of juice did she buy

kupik [55]

You multiply by 6

64 * 6 = 384