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Nookie1986 [14]
3 years ago
7

A rectangle with 8 cm width and 3 cm length What is the perimeter and area of the rectangle?​

Mathematics
1 answer:
kompoz [17]3 years ago
4 0

Answer:

P=22cm

A=24cm²

Step-by-step explanation:

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Boa saved $179 a month He saved $145 less than Ada each month. how much would Ada save in three and a half years
Licemer1 [7]
  I think your answer would be $69984. But I'm not 100 percent.
5 0
3 years ago
Find the range of the following piecewise function. [27,3) (-22,3] (27,3] [-22,3)
Otrada [13]

Answer:

The answer is "(27,3]"

Step-by-step explanation:

Please find the complete question in the attached file.

\to f(-5) = (-5)^2 + 2 = 25 +2 =27\\\\\to f(7)= 7-4 = 3\\\\answer \ is = ( 27, 3]\

5 0
3 years ago
Read 2 more answers
Show that the two triangles ABC and DEF are congruent by the<br> HL theorem.
inna [77]

Since hypotenuses and legs have the same length, then both are congruent and thus, triangles ABC and DEF are congruent.

The Hypotenuse-Leg theorem states that two <em>right</em> triangles are <em>congruent</em> if and only if their hypotenuses and any of their <em>corresponding</em> legs are <em>congruent</em>. That is to say:

\overline {AC} \cong \overline {DF}, \overline{BC} \cong \overline {EF}

AC = DF, BC = EF

\sqrt{AB^{2}+BC^{2}} = \sqrt{DE^{2}+EF^{2}}, BC = EF

If we know that AB = 10, BC = 4, DE = 10 and EF = 4, then we have the following outcomes:

\sqrt{10^{2}+4^{2}} = \sqrt{10^{2}+4^{2}}

4 = 4

Since hypotenuses and legs have the same length, then both are congruent and thus, triangles ABC and DEF are congruent.

We kindly invite to check this question on triangles: brainly.com/question/21972776

3 0
3 years ago
On Monday, students at a summer camp spent 4 hours 25 minutes at the pool learning to swim. In the morning, they spent 2 hours 4
Lilit [14]
4 hours and 25 minutes simplifies to 265 minutes
2 hours and 48 minutes simplifies to 168 minutes

Subtract the numbers: 265-168 = 97 minutes left

Convert back to hours and you have 1 Hour 37 minutes at the pool in the afternoon.
5 0
3 years ago
The weekly amount of money spent on maintenance and repairs by a company was observed, over a long period of time, to be approxi
VikaD [51]

Answer:

$512.90 should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.05

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Normally distributed with mean $480 and standard deviation $20.

This means that \mu = 480, \sigma = 20

How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.05?

This is the 100 - 5 = 95th percentile, which is X when Z has a pvalue of 0.95, so X when Z = 1.645.

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

1.645 = \frac{X - 480}{20}

X - 480 = 1.645*20

X = 512.9

$512.90 should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.05

5 0
3 years ago
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