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Dmitry_Shevchenko [17]
2 years ago
9

Help me please

Mathematics
1 answer:
Fittoniya [83]2 years ago
8 0
110a3b3c4 is yours answer.
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I need the answers for these pls help
DanielleElmas [232]

Answer:

1)

A. m arc QRS = 100°

B. m arc QRT = 255°

C. m arc UTS = 180°

D. m arc UTR = 205°

2)

1] The major arc CA = 255°

2] The minor arc AB = 90°

Step-by-step explanation:

In a circle:

  • The measure of an central angle is equal to the measure of its subtended arc
  • The measure of a circle is 360°
  • Any chord divides a circle into two arcs, a minor arc its measure is < 180° and a major arc its measure is > 180°, the sum of the measures of the minor and major arcs is 360°
  • If the measures of the minor and major arcs are equal, then each arc represents a semi-circle

1)

In circle O

A.

∵ m∠QOR = 75°

∵ ∠QOR subtended by arc QR

- By using the 1st note above

∴ m of arc QR = m∠QOR

∴ m of arc QR = 75°

∵ m∠ROS = 25°

∵ ∠ROS subtended by arc RS

∴ m of arc RS = m∠ROS

∴ m of arc RS = 25°

The measure of arc QRS is the sum of the measures of arcs QR and RS

∴ m arc QRS = 75° + 25° = 100°

B.

∵ m∠SOT = 155°

∵ ∠SOT subtended by arc ST

∴ m of arc ST = m∠SOT

∴ m of arc ST = 155°

The measure of arc QRT is the sum of the measures of arcs QR, RS and ST

∴ m arc QRT = 75° + 25° + 155 °= 255°

C.

∵ m∠UOT = 25°

∵ ∠UOT subtended by arc UT

∴ m of arc UT = m∠UOT

∴ m of arc RUT = 25°

The measure of arc UTS is the sum of the measures of arcs UT and TS

∴ m arc UTS = 25° + 155° = 180°

D.

The measure of arc UTR is the sum of the measures of arcs UT, TS and SR

∴ m arc UTR = 25° + 155° + 25 = 205°

2)

1]

∵ The sum of the measures of the minor and major arcs is 360°

∵ m of minor arc AC = 105°

- Subtract 105 from 360° to find the measure of the major arc CA

∴ m of major arc CA = 360° - 105°

∴ m of major arc CA = 255°

2]

∵ m of major arc AB = 270°

- Subtract 270° from 360° to find the measure of the minor arc AB

∴ m of minor arc AB = 360° - 270°

∴ m of minor arc AB = 90°

8 0
3 years ago
Did i do this correctly?
den301095 [7]
Yes u did do this correctly
5 0
3 years ago
Read 2 more answers
Circle all the words that describe the quadrilateral
svp [43]
Rombus I guess I don’t really know
3 0
3 years ago
Read 2 more answers
Construct a 90% confidence interval for μ1-μ2 with the sample statistics for mean calorie content of two​ bakeries' specialty pi
DIA [1.3K]

Answer:

The 90% confidence interval for the difference in mean (μ₁ - μ₂) for the two bakeries is; (<u>49</u>) < μ₁ - μ₂ < (<u>289)</u>

Step-by-step explanation:

The given data are;

Bakery A

\overline x_1<em> </em>= 1,880 cal

s₁ = 148 cal

n₁ = 10

Bakery B

\overline x_2<em> </em>= 1,711 cal

s₂ = 192 cal

n₂ = 10

\left (\bar{x}_1-\bar{x}_{2}  \right ) - t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}< \mu _{1}-\mu _{2}< \left (\bar{x}_1-\bar{x}_{2}  \right ) + t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}

df = n₁ + n₂ - 2

∴ df = 10 + 18 - 2 = 26

From the t-table, we have, for two tails, t_c = 1.706

\hat{\sigma} =\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}

\hat{\sigma} =\sqrt{\dfrac{\left ( 10-1 \right )\cdot 148^{2} +\left ( 18-1 \right )\cdot 192^{2}}{10+18-2}}= 178.004321469

\hat \sigma ≈ 178

Therefore, we get;

\left (1,880-1,711  \right ) - 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}< \mu _{1}-\mu _{2}< \left (1,880-1,711  \right ) + 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}

Which gives;

169 - \dfrac{75917\cdot \sqrt{35} }{3,750} < \mu _{1}-\mu _{2}< 169 + \dfrac{75917\cdot \sqrt{35} }{3,750}

Therefore, by rounding to the nearest integer, we have;

The 90% C.I. ≈ 49 < μ₁ - μ₂ < 289

4 0
3 years ago
Simplify the expression. <br> Can someone explain to me how to get the answer? Thanks!
Lynna [10]

Answer:

11x^2+2x+9

Step-by-step explanation:

2+(2+6)⋅x^2+2x+3x^2+7

2+(8)⋅x^2+2x+3x^2+7

2+8x^2+2x+3x^2+7

9+8x^2+2x+3x^2

9+8x^2+2x+3x^2

11x^2+2x+9

3 0
3 years ago
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