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maks197457 [2]
2 years ago
11

Aaron drew the figure below for a school art project. What is the total area of the figure. Use the drop down menus to complete

the sentences to determine the total area of the figure. Please help I need it soon

Mathematics
2 answers:
Anna11 [10]2 years ago
7 0

Answer:

Step-by-step explanation:

Feliz [49]2 years ago
6 0

Answer:

The trapezoid has a height of 3 inches, a shorter base measuring 2.75 inches, and a longer base measuring 4.25 inches. The total area of the entire figure is 33 square inches.

Step-by-step explanation:

I took the iready diagnostic :)

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Sammie bought just enough fencing to border either a rectangular plot or a square plot, as shown. The perimeters of the plots ar
Gekata [30.6K]
Perimeter of square : p = 4a.....
p = 4(x + 2)
p = 4x + 8

perimeter of rectangle : p = 2(L + W)
p = 2(3x + 2 + x - 1)
p = 2(4x + 1)
p = 8x + 2

so if the perimeters are te same, lets set them equal to each other and solve for x

4x + 8 = 8x + 2
8 - 2 = 8x - 4x
6 = 4x
6/4 = x
1.5 = x

the square : p = 4x + 8.....p = 4(1.5) + 8.....p = 14 meters
the rectangle : p = 8x + 2....p = 8(1.5) + 2.....p = 14 meters

so she bought 14 meters of fencing <==
8 0
2 years ago
Select all the equations where n= 6 is a solution.
timurjin [86]

Answer:

4=10-n

3=n/2

8n=48

Step-by-step explanation:

Because for the first one instead of subtracting 10-n you will subtract 10-4, which is 6 so that one and for the second one you would do the inverse of division and multiply 3x2and get 6 and then for the fourth one you will divide 8/48 and get 6

Hoped this helped!!! Have a Great Day!!! Pls give me BRAINLIEST!!

8 0
2 years ago
Read 2 more answers
The temperature reading from a thermocouple placed in a constant-temperature medium is normally distributed with mean μ, the act
stealth61 [152]

Answer:  0.2551

Step-by-step explanation:

Given : The temperature reading from a thermocouple placed in a constant-temperature medium is normally distributed with mean μ, the actual temperature of the medium, and standard deviation σ.

Significance level : \alpha=1-0.95=0.05

The critical z-value for 95% confidence : z_{\alpha/2}=1.960 (1)

Since , z=\dfrac{x-\mu}{\sigma} (where x be any random variable that represents the temperature reading from a thermocouple.)

Then, from (1)

\dfrac{x-\mu}{\sigma}=1.96\\\\ x-\mu=1.96\sigma     (2)

Also,  all readings are within 0.5° of μ,

i.e. x-\mu

i.e. 1.96\sigma   [From (2)]

i.e. \sigma  

i.e. \sigma\approx0.2551

The required standard deviation : \sigma=0.2551

3 0
2 years ago
There are exactly π radians in a semicircle. There are exactly π radians in a full circle. There are approximately radians in a
Allisa [31]

pi radians in a semicircle (half circle)

2pi radians in a full circle.

2pi = 2*3.14 = 6.28 approximately, so there are approximately 6.28 radians in a half circle.

2pi radians is equivalent to 360 degrees.


8 0
2 years ago
Read 2 more answers
A mile-runner’s times for the mile are normally distributed with a mean of 4 min. 3 sec. (This would have to be expressed in dec
Vinvika [58]

Answer: 0.0668

Step-by-step explanation:

Given: Mean : \mu=\text{4 min. 3 sec.=4.05 minutes}

Standard deviation : \sigma = \text{2 seconds=0.033333 minutes }

The formula to calculate z-score is given by :_

z=\dfrac{x-\mu}{\sigma}

For x= 4 minutes  , we have

z=\dfrac{4-4.05}{0.03333}\approx-1.5

The P-value = P(z\leq-1.5)=0.0668072\approx0.0668

Hence, the probability that on a given run, the time will be 4 minutes or less = 0.0668

8 0
2 years ago
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