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RSB [31]
3 years ago
14

A rectangle is inside a circle with a 5 cm radius.

Mathematics
1 answer:
MrRa [10]3 years ago
3 0

Answer:  30.5 cm²

To get the the area of the rectangle all we have to do is:

Rectangle Area = 6 × 8 = 48 cm²

Same thing to the circle, multiply the radius and pi:

Circle Area = 3.14 x 5² = 78.5 cm²

We have to subtract the two areas to find the shaded area.

78.5 - 48 = 30.5 cm²

I hope this helps!

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Did I get this first question right?
vivado [14]

Answer:

no

Step-by-step explanation:

the correct answer is 56.55 inches

3 0
3 years ago
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Question Help Suppose that the lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a
koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean \mu and standard deviation \sigma, the z-score of a value X is given by:

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

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So \mu = 5656, \sigma = 333.3

(a) What proportion of light bulbs will last more than 6262 ​hours?

The pvalue of the z-score of X = 6262 is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.

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

Z = \frac{6262 - 5656}{333.3}

Z = 1.82

Z = 1.81 has a pvalue of .96562. This means that 96.562% of the light bulbs are going to last less than 6262 hours. So

P = 100% - 96.562% = 3.438% of the light bulbs will last more than 6262 hours.

​(b) What proportion of light bulbs will last 5252 hours or​ less?

This is the pvalue of the zscore of X = 5252

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

Z = \frac{5252- 5656}{333.3}

Z = -1.21

Z = -1.21 has a pvalue of .1131. This means that 11.31% of the light bulbs will last 5252 hours or less.

(c) What proportion of light bulbs will last between 5858 and 6262 ​hours?

This is the pvalue of the zscore of X = 6262 subtracted by the pvalue of the zscore X = 5858

For X = 6262, we have that Z = 1.81 with a pvalue of .96562.

For X = 5858

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

Z = \frac{5858- 5656}{333.3}

Z = 0.61

Z = 0.61 has a pvalue of .72907.

So, the proportion of light bulbs that will last between 5858 and 6262 hours is

P = .96562 - .72907 = .23655

23.655% of the light bulbs are going to last between 5858 and 6262 hours.

​(d) What is the probability that a randomly selected light bulb lasts less than 4646 ​hours?

This is the pvalue of the zscore of X = 4646

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

Z = \frac{4646- 5656}{333.3}

Z = -3.03

Z = -3.03 has a pvalue of .0012. This means that 0.12% of the light bulbs will last 4646 hours or less.

5 0
3 years ago
9. Evaluate 14(-6x +10) - 16x - 100x + 140 for x = -2.
Radda [10]

Answer:

680

Step-by-step explanation:

14(-6x +10) - 16x - 100x + 140 for x = -2.

14(-6(-2) + 10) - 16(-2) - 100(-2) + 140

14(12 + 10) + 32 + 200 + 140

14(22) + 372

308 + 372

680

5 0
2 years ago
Evaluate (212)base3-(121)base3+(222)base3​
NemiM [27]

Answer:

212 bp^3 (base pairs cubed) - 121 bp^3 (base pairs cubed) + 222 bp^3 (base pairs cubed) = 313 bp^3 (base pairs cubed)

Step-by-step explanation:

3 0
3 years ago
X/2+3= 5 its solving equations (with three terms) please help me
olganol [36]
Multiply each side by 2: 2(x/2) + 3(2) = 5(2) => x + 6 = 10

Subtract 6 from each side: x + (6 - 6) = 10 - 6 => x = 4

The answer is x is equal to 4.
8 0
3 years ago
Read 2 more answers
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