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bearhunter [10]
2 years ago
11

Find the volume of the figure if length = 1/2 m, width = 1/2 m and height = 6 meters. I WILL GIVE BRAINLIEST PLLZZZ SOMEONE HELP

ME
1/24 m cubed

1 1/2 m cubed

3 m cubed

7 m cubed
Mathematics
1 answer:
goldfiish [28.3K]2 years ago
5 0

Step-by-step explanation:

Volume of figure=length×width×height

=(1/2)×(1/2)×6

=0.5×3

=1.5m³

PLEASE GIVE BRAINLIEST

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stealth61 [152]

Apply law of total resistance in parallel circuit

\\ \tt\hookrightarrow \dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}

\\ \tt\hookrightarrow \dfrac{1}{9+2J}=\dfrac{1}{3-5j}+\dfrac{1}{R_2}

\\ \tt\hookrightarrow \dfrac{1}{R_2}=\dfrac{1}{9+2J}-\dfrac{1}{3-5J}

\\ \tt\hookrightarrow \dfrac{1}{R_2}=\dfrac{3-5J-9-2J}{(9+2J)(3-5J)}

\\ \tt\hookrightarrow R_2=\dfrac{(9+2J)(3-5J)}{-6-7J}

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2 years ago
What is 5+3*4-8+2*7 equal to
neonofarm [45]

Answer: 23 is your answer


Step-by-step explanation:


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Your history teachers is going to give you a 92 point assessment. There are 2 point questions and 3 point questions. If there is
Temka [501]

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28

Step-by-step explanation:

x = 2 point questions, y = 3 point questions

x + y = 40

2x + 3y = 92

Solving this we get x = 28, y = 12 so the answer is 28.

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A survey found that women's heights are normally distributed with mean 62.7 in, and standard deviation 2.8 in. The survey also f
Viefleur [7K]

Using the normal distribution, we have that:

The percentage of men who meet the height requirement is 3.06%. This suggests that the majority of employees at the park are females.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation of men's heights are given as follows:

\mu = 69.3, \sigma = 3.9.

The proportion of men who meet the height requirement is is the <u>p-value of Z when X = 62 subtracted by the p-value of Z when X = 55</u>, hence:

X = 62:

Z = \frac{X - \mu}{\sigma}

Z = \frac{62 - 69.3}{3.9}

Z = -1.87

Z = -1.87 has a p-value of 0.0307.

X = 55:

Z = \frac{X - \mu}{\sigma}

Z = \frac{55 - 69.3}{3.9}

Z = -3.67

Z = -3.67 has a p-value of 0.0001.

0.0307 - 0.0001 = 0.0306 = 3.06%.

The percentage of men who meet the height requirement is 3.06%. This suggests that the majority of employees at the park are females.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

8 0
1 year ago
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