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dedylja [7]
4 years ago
6

The standard error of the estimate measures the scatter or dispersion of the observed values around a __________________________

________________________________
Mathematics
1 answer:
muminat4 years ago
7 0

Answer:

True mean/population mean

Step-by-step explanation:

The standard error in this case gives an estimate on how far the values observed during the course of the experiment ate likely to be from the true mean/population mean.

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15. Scott and Tara were asked to find the
Dmitriy789 [7]

Answer:

Both Scott and Tara have responded correctly.

Step-by-step explanation:

we know that

The area of a trapezoid is equal to

A=(1/2)[b1+b2]h

we have

b1=16 cm

b2=24 cm

h=8 cm -----> <em>Note</em> The height is 8 cm instead of 18 cm

substitute

A=(1/2)[16+24](8)

A=160 cm²

<em>Verify Scott 's work</em>

<em>Note</em> Scott wrote A = (1/2)(24 + 16)(8) instead of A = 2(24 + 16)(8)

Remember that the Commutative Property establishes "The order of the addends does not alter its result"

so

(24+16)=(16+24)

A = (1/2)(24 + 16)(8)=160 cm²

<em>Verify Tara's work</em>

<em>Note</em> Tara wrote A = (1/2)(16+24)(8) instead of A = (16 + 24)(8)

A = (1/2)(16+24)(8)=160 cm²

5 0
3 years ago
Brady made a scale drawing of a rectangular swimming pool on a coordinate grid. The points (-20, 25), (30, 25), (30, -10) and (-
djverab [1.8K]

Answer:

Length = 50 units

width = 35 units

Step-by-step explanation:

Let A, B, C and D be the corner of the pools.

Given:

The points of the corners are.

A(x_{1}, y_{1}})=(-20, 25)

B(x_{2}, y_{2}})=(30, 25)

C(x_{3}, y_{3}})=(30, -10)

D(x_{4}, y_{4}})=(-20, -10)

We need to find the dimension of the pools.

Solution:

Using distance formula of the two points.

d(A,B)=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}----------(1)

For point AB

Substitute points A(30, 25) and B(30, 25) in above equation.

AB=\sqrt{(30-(-20))^{2}+(25-25)^{2}}

AB=\sqrt{(30+20)^{2}}

AB=\sqrt{(50)^{2}

AB = 50 units

Similarly for point BC

Substitute points B(-20, 25) and C(30, -10) in equation 1.

d(B,C)=\sqrt{(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}}

BC=\sqrt{(30-30)^{2}+((-10)-25)^{2}}

BC=\sqrt{(-35)^{2}}

BC = 35 units

Similarly for point DC

Substitute points D(-20, -10) and C(30, -10) in equation 1.

d(D,C)=\sqrt{(x_{3}-x_{4})^{2}+(y_{3}-y_{4})^{2}}

DC=\sqrt{(30-(-20))^{2}+(-10-(-10))^{2}}

DC=\sqrt{(30+20)^{2}}

DC=\sqrt{(50)^{2}}

DC = 50 units

Similarly for segment AD

Substitute points A(-20, 25) and D(-20, -10) in equation 1.

d(A,D)=\sqrt{(x_{4}-x_{1})^{2}+(y_{4}-y_{1})^{2}}

AD=\sqrt{(-20-(-20))^{2}+(-10-25)^{2}}

AD=\sqrt{(-20+20)^{2}+(-35)^{2}}

AD=\sqrt{(-35)^{2}}

AD = 35 units

Therefore, the dimension of the rectangular swimming pool are.

Length = 50 units

width = 35 units

7 0
3 years ago
The relationship between the amount of data downloaded d, in megabytes, and the time t, in seconds, after the download started i
svetoff [14.1K]

Answer: A, C, F, there’s 4 but I only know 3

6 0
3 years ago
Read 2 more answers
Round 1,490 to the nearest hundred
Arlecino [84]
The answer is 1,500..
3 0
3 years ago
Read 2 more answers
Find the area of this regular polygon.<br> Round to the nearest tenth.<br> 8.65 mm<br> [? ]mm2
nadezda [96]

Answer:

Actually it's not polygon. it's a nonagon. With r=8.65mm″, the law of cosines gives us side a:

a=√{b²+c²−2bc×cos40°}

a=√{149.645−149.645cos40°}

Area Nonagon = (9/4)a²cos40°

=9/4[149.645−149.645cos40°]cot20°

=336.70125[1−cos(40°)]cot(20°)

Applying an identity for the cos(40°) does not get us very far…

= 336.70125[1−(cos2(20°)−1)]cot(20°)

= 336.70125[2−cos2(20°)]cot(20°)

= 336.70125[2−(1−sin2(20°))]cot(20°)

= 336.70125[1+sin2(20°)]cos(20°)sin(20°)

= 336.70125[cot(20°)+sin(20°)cos(20°)]mm²

3 0
3 years ago
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