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2 years ago

Which image shows a plane slicing a double-napped cone to produce a hyperbola? Answer is D

Mathematics

1 answer:

Butoxors [25]2 years ago

4 0

The <u>second image</u> in the diagram is a hyperbola. As can be seen, the plane cutting the cone can be at any angle but never equal to the slant angle of the cone. This has a very important implication. The plane cuts both cones of the double-napped cone. The third double-napped cone of Figure 3 shows two hyperbolas.

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Find the measure of the exterior angle. 86° 46° xº.

Anna11 [10]

Answer:

Step-by-step explanation:

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3 years ago

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Arlecino [84]

25% is equal to 1/4 so 180 times 3/4 equals $135

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3 years ago

dolores and four friends went to a buffet dinner.The total cost was at most $130 including the $20 dollar tip they left.How much

erma4kov [3.2K]

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3 years ago

Read 2 more answers

Multiply.<br><br> (2.1×102)×(3.5×105)

hichkok12 [17]

Answer:

78,718.5

Step-by-step explanation:

2.1*102=214.2

3.5*105=367.5

214.2*367.5=78,718.5

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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