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vesna_86 [32]
3 years ago
12

What is the relationship between segment ED and segment AC?

Mathematics
1 answer:
Bas_tet [7]3 years ago
6 0
D midpoint of EC -----------------> FD parallel to AC and FD=AC/2=14/2=7 
<span>2-EB=EA E midpoint of AB </span>
<span>DB=DC D midpoint BC ...............> ED=AC/2=2 </span>
<span>3-T midpoint of SR </span>
<span>U midpoint of QR ---------> TU = QS/2 </span>
<span>QS=2 TU = 4.4 </span>
<span>4- The same steps SR=2 UV=9 </span>
<span>5-N midpoint of KM </span>
<span>O midpoint of ML </span>
<span>* NO parallel to Kl</span>
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A 99% confidence interval for the sugar content (per 5 oz. serving) of a shipment of strawberries is 5.9 to 8.3 grams.
gavmur [86]

Answer:

Sample mean = 7.1

Margin of error = 0.465

Step-by-step explanation:

Formula for confidence interval is;

CI = x¯ ± zE

Where;

x¯ is sample mean

z is critical value at confidence level

E is margin of error.

z for 99% Cl is 2.58

We are told the CI is 5.9 to 8.3.

Thus;

5.9 = x¯ - 2.58E - - - - (1)

8.3 = x¯ + 2.58E - - - - (2)

Add both equations together to get;

14.2 = 2x¯

x¯ = 14.2/2

x¯ = 7.1

Put 7.1 for x¯ in eq 1 to get;

5.9 = 7.1 - 2.58E

7.1 - 5.9 = 2.58E

E = 1.2/2.58

E = 0.465

5 0
3 years ago
List the factors of 24.
Sergio [31]

Answer:

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Step-by-step explanation:

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7 0
3 years ago
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Uranus moves in an elliptical orbit with the sun at one of the foci. The length of the half of the major axis is 2,876,769,540 k
Alina [70]

Answer:

The minimum distance (perihelion) of Uranus from the sun is 2,749,040,972.

Step-by-step explanation:

Consider the provided information.

The length of the half of the major axis is 2,876,769,540 kilometers, and the eccentricity is 0.0444.

The eccentricity (e) of an ellipse is the ratio of the distance from the center to the foci (c) and the distance from the center to the vertices (a).

e=\frac{c}{a}

Substitute a = 2,876,769,540 and e = 0.0444 in above formula and solve for c.

0.0444=\frac{c}{2,876,769,540 }

c=127728567.576

Minimum distance of Uranus from the sun is:

a-c=2,876,769,540-127728567.576\\a-c=2749040972.424\approx2749040972

Hence, the minimum distance (perihelion) of Uranus from the sun is 2,749,040,972.

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3 years ago
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mixas84 [53]

Answer:

Step-by-step explanation:

b x 1

6 0
2 years ago
Answer quickly for brainliest and 5 star!
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