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3 years ago

What is 1.5 to the third power?

1 answer:

7 0

3.375 would be the answer, exponents with decimals is the same as regular exponents, just multiple the base number times the exponent you are working with

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Marko is playing the video game Fort Attack. The purpose of the game is to shoot invading bandits that are trying to breach the

shutvik [7]

Angle of fire in degrees: 90° + 30° = 120°

Angle of fire in radians: 180° = π

120° * π/180° = 120°π / 180° OR 4π/6 OR 2π/3

4 0

3 years ago

Read 2 more answers

I GIVES BRAINLIEST TO PEOPLES WHO ANSWERS

CaHeK987 [17]

Answer:

A) $28.14 B) $8.14

Step-by-step explanation:

1 * 12.95 = $12.95

2 * 3.20 = $6.40

1 * 8.79 = $8.79

we add up all these numbers, 28.14

well he has only 20 bucks, so he still needs 28.14-20= $8.14

Please don't forget to give brainliest ;)

7 0

3 years ago

Read 2 more answers

IN DESPERATE NEED OF HELP!!!! <br> WILL MARK BRAINLIEST!!

White raven [17]

For the squirrel questions
1.c
2.b
3.d
4.a

7 0

3 years ago

Read 2 more answers

X-y=10 in the form of y=mx+b

creativ13 [48]

Y = 10+x
(Moving the x)

y = x+10
(Reordering the x so it becomes y=mx+b)

The equation is y = x + 10

Hope this helps! :)

4 0

4 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

8 0

3 years ago