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2 years ago

A four sided sandbox has exactly 2 right angles to side lengths 5 feet what geometric shape as described shape of the sandbox

Mathematics

1 answer:

lana66690 [7]2 years ago

3 0

Answer:

The sandbox is most likely in the shape of a kite. As it has only two right angles, it cannot be a square or other form of rectangle. It also cannot be a parallelogram, which has no right angles, nor is it a rhombus- all four sides are not of the same length.

A kite, however, has two pairs of equal-length sides that are adjacent to each other. It also has two sets of equal-measure angles, one of which can be 90° if it is a cyclic kite.

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Determine the measure of EBD

Soloha48 [4]

Answer:

Answer:

EBD= 156

Step-by-step explanation:

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3 years ago

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During a period of 30 working days, Deborah was late 80% of the time. How many days was she late?

erik [133]

Fairly straightforward question. She works 30 days and is late 80% or 0.8
We can model our equation
Num days late = late percentage * num working days
Basically fine portion of working days that she was late.

Num days late = 0.8 * 30=24

So Deborah was late 24 out of 30 working days. Would appreciate a brainliest

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2 years ago

-4y+8-3x+4y+9+5x= Need help

lord [1]

Answer:

2x+17

Step-by-step explanation:

-4y+8-3x+4y+9+5x

Combine like terms

-3x +5x -4y +4y +8+9

2x +0 +17

7 0

3 years ago

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Assume that 33.4% of people have sleepwalked. Assume that in a random sample of 1459 adults, 526 have sleepwalked.

lutik1710 [3]

Answer:

Step-by-step explanation:

Given that 33.4% of people have sleepwalked.

Sample size n =1459

Sample favourable persons = 526

Sample proportion p = $\frac{526}{1459} \\=0.361$

Sample proportion p is normal for large samples with mean = 0.334 and

std error = $\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.334(1-0.334)}{1459} } \\=0.0123$

a) P(526 or more of the 1459 adults have sleepwalked.)

$=P(p\geq 0.361)\\=P(Z\geq \frac{0.361-0.334}{0.0123} \\=P(Z\geq 2.20)\\=0.5-0.4861\\=0.0139$

b) Yes, because hardly 1.4% is the probability

c) 33.4 is very less compared to the average. Either sample should be improved representing the population or population mean should be increased accordingly.

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3 years ago

The Academic Computing Center has five trainers available in its computer labs to provide training sessions to students. Assume

kondor19780726 [428]

Answer:

87.72%

Step-by-step explanation:

Data provided in the question:

Design capacity of the system = 1900 students per semester

Effective capacity = 90% of design capacity

Actual number of students = 1500

Now,

Efficiency = [ [ Actual capacity ] ÷ [ Effective capacity ] ] × 100%

also,

Effective capacity = 90% of 1900

= 0.90 × 1900

= 1710

Efficiency = [ 1500 ÷ 1710 ] × 100%

= 0.8772 × 100%

= 87.72%

7 0

3 years ago