All categories

3 years ago

How to answer number 10

Mathematics

1 answer:

Ipatiy [6.2K]3 years ago

5 0

You were on the right track.

(17/4)*(11/4)*(3/1)

first multiply the numerators

17*11*3 = 561

561 / ?

next multiply the denominators

4*4*1 = 16

561 / 16 = 35.0625 = 35 1/16

You might be interested in

Please answer I don’t understand what neatest integer means ??

umka21 [38]

Idk but good louky nad happy new year

4 0

3 years ago

Read 2 more answers

Help me with this quick pleaseeeee

melisa1 [442]

Answer:

D' = ( 1 , -1 )

Step-by-step explanation:

2 times 1/2 = 1

-2 times 1/2 = -1

hope this helps ^^

3 0

3 years ago

Right triangle ABC is shown below.

Flura [38]

The answer is 12

Step-by-step explanat

5 0

4 years ago

Please help me with this question.... please...

ivanzaharov [21]

Answer:

just plug in the x value into the equation and put it in ur calculator

3 0

3 years ago

Read 2 more answers

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