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2 years ago

Mrs.Gunn makes one costume in 3hours and 20 minutes. How long will it take her to make 6 costumes for a school play?

Mathematics

2 answers:

qaws [65]2 years ago

4 0

Answer:

20 hours

Step-by-step explanation:

Tpy6a [65]2 years ago

4 0

We can change the 3 hours and 20 minutes to "minutes" instead of hours:

1 hour is 60 minutes. So 3 hours is 60 * 3 = 180 minutes.

Add the 20 : 180 + 20 = 200 minutes for one costume.

Multiply by 6 for 6 costumes: 200 * 6 = 1200.

So 1200 minutes or 1200/60 = 20 hours to make 6 costumes

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There are 81 red, blue and yellow

wel

Step-by-step explanation:

Here Given

Total counters = 81

red counters are 9 more than the yellow ones.

And Yellow counter = blue counter

So let the Yellow counter be x and as blue counter and yellow counter are same then blue counter will be also x

And by the question

Red counter = 9 + x

Now

9 + x + x + x = 81

9 + 3x = 81

3x = 81 - 9

3x = 72

x = 72 / 3

Therefore x = 24

Now

Red counters = 9 + 24 = 33

Blue counters = 24

Yellow counters = 24

Hope it helped :)

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2 years ago

Read 2 more answers

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Scilla [17]

Answer:

$\checkmark \text{B. The area of the circle is } 729\pi,\\\checkmark \text{D. The arc length of the sector is } 12\pi$

Step-by-step explanation:

Area of a circle with radius $r$ : $r^2\pi$

Circumference of a sector with radius $r$ : $2r\pi$

Area of sector with angle $\theta$ : $r^2\pi\cdot \frac{\theta}{360}$ .

Arc length of sector with angle $\theta$ : $2r\pi\cdot \frac{\theta}{360}$

Using these equations, we get the following information:

Area of circle: $27^2\pi=729\pi$

Circumference of a circle: $2\cdot 27\cdot \pi=54\pi$

Area of sector: $27^2\pi\cdot \frac{80}{360}=162\pi$

Arc length of sector: $2\cdot 27\cdot \pi \cdot \frac{80}{360}=12\pi$

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2 years ago

Giving brainliest!!!!!!! Please be correct if you don’t know the answer don’t andwer

mars1129 [50]

Answer:

Step-by-step explanation:hope this helps have a great day =D

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2 years ago

In Nebraska, the mean rain-fed corn yield of harvested fields are normally

NikAS [45]

Answer: 169.2 per acre are produced if a field is in the top 25%.

Step-by-step explanation:

Given: The mean rain-fed corn yield of harvested fields are normally distributed, produce 147.2 bushels per acre with a standard deviation of 33 bushels per acre.

Let X denote the mean rain-fed corn yield of harvested fields.

Let x denotes the number of bushels per acre are produced if a field is in the top 25%.

Top 25% on a density curve means , (100-25%)i.e. 75% area lies below x.

I.e.

$P(X$ .

Z-score for p-value of 0.75 (one tailed)=0.67 [By z-table]

Then, $\dfrac{x-147.2}{33}=0.6744$

$\Rightarrow\ {x-147.2}=33\times0.67\\\\\Rightarrow\ {x-147.2}=22.11\\\\\Rightarrow\ {x}=22.11+147.2=169.31$

Here approx value is 169.2 (from options).

So, 169.2 per acre are produced if a field is in the top 25%.

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3 years ago

A high school athletic department bought 40 soccer uniforms at a cost of$3,000. after soccer season, they returned some of the u

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75, division is key for this problem

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