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2 years ago

5

How much soup, to the nearest tenth of a cubic unit, can fit in a cylindrical can that is 8 inches tall and has a diameter of 6

inches?

1 answer:

Dovator [93]2 years ago

8 0

Answer:

Step-by-step explanation:

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two planes take off at a new york airport. the first plane climbs at an angle of 45 degree with respect to the ground amd reache

vaieri [72.5K]

The answer is fully given in the picture below. Hope it helps!

6 0

2 years ago

Is this question correct?

Andrews [41]

Answer:

10^20

3^11

3^24

10^12

Step-by-step explanation:

(10^2)^10=10^2*10=10^20

3^3*3^8=3^3+8=3^11

(3^3)^8=3^3*8=3^24

10^2*10^10=10^2+10=10^12

6 0

2 years ago

In a bolt factory, machines 1,2, and 3 respectively produce 20%, 30% and 50% of the total output. of their respective outputs 5%

lorasvet [3.4K]

Answer:

a. 1/10 or 10%

b. 1/2 or 50%

Step-by-step explanation:

Since the combination of machines 1, 2 and 3 produce 100% of the total output when added together, then the probability of choosing a bolt at random that is defective is: 5 + 2 + 3 = 10% out of 100% or 10/100, which is 1/10 or 10%.

If the bolt that is choosen at random is defective, than the probability that it came from machine 1 is 5/10 or 1/2 which is also 50%.

3 0

3 years ago

Fill in the blank (no word bank)

timofeeve [1]

Answer:

Alternate Interior Angles:

∠3 ≅ ∠5

Corresponding Angles:

∠3 ≅ ∠7

∠4 ≅ ∠8

Supplementary Angles:

∠3 is supplementary to ∠6

Step-by-step explanation:

7 0

2 years ago

A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normall

Vlad1618 [11]

Answer: (2.54,6.86)

Step-by-step explanation:

Given : A random sample of 10 parking meters in a beach community showed the following incomes for a day.

We assume the incomes are normally distributed.

Mean income : $\mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7$

Standard deviation : $\sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}$

$=\sqrt{\dfrac{(1.1)^2+(0.2)^2+(1.9)^2+(1.6)^2+(2.1)^2+(0.5)^2+(2.05)^2+(0.45)^2+(3.3)^2+(1.7)^2}{10}}$

$=\dfrac{30.265}{10}=3.0265$

The confidence interval for the population mean (for sample size <30) is given by :-

$\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given significance level : $\alpha=1-0.95=0.05$

Critical value : $t_{n-1,\alpha/2}=t_{9,0.025}=2.262$

We assume that the population is normally distributed.

Now, the 95% confidence interval for the true mean will be :-

$4.7\ \pm\ 2.262\times\dfrac{3.0265}{\sqrt{10}} \\\\\approx4.7\pm2.16=(4.7-2.16\ ,\ 4.7+2.16)=(2.54,\ 6.86)$

Hence, 95% confidence interval for the true mean= (2.54,6.86)

7 0

2 years ago