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ziro4ka [17]
3 years ago
6

Please help I'll give brainliest

Mathematics
1 answer:
devlian [24]3 years ago
6 0

Answer:

m∠1 + m∠3

Step-by-step explanation:

Please please mark me brainliest and have a great day!

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What is x equal to in the problem 10+3(x+2)=36? Please show your work.
scZoUnD [109]

Answer:

x = 6 2/3

Step-by-step explanation:

10+3(x+2)=36

10+3x+6=36           Solve inside the parenthesis.

16+3x=36               Combine like terms.

-16       -16                Subtract 16 from both sides.

3x/3=20/3                Divide both sides by 3.

x=6 2/3                     The answer is x= 6 2/3, or 6.66666666666667

5 0
3 years ago
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Stephanie saw an ad in the paper that her favorite kind of perfume was on sale for 25% off. Her favorite perfume costs $20 per b
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Answer:

$5

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3 years ago
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A ribbon measures <br> 4.5 meters. Express this length in millimeters.
Veseljchak [2.6K]

Answer:

4500 meters.

Step-by-step explanation:

1 meter = 1000 millimeters so just do the number of meters (in this case 4.5) times 1000.

8 0
3 years ago
Heather flipped a coin five times, and each time it came up heads. Heather flips the coin one more time. Which is the theoretica
makkiz [27]

Answer:

1/2

Step-by-step explanation:

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4 years ago
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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
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