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3 years ago

Find the unknown angle

Mathematics

2 answers:

Naily [24]3 years ago

8 0

Answer:

Step-by-step explanation:

X=180_(90+41)=49

leonid [27]3 years ago

8 0

Answer:

x = 49

Step-by-step explanation:

180 = 41 + 90 + x (90 bc it a rt angle btw)

180 = 131 + x

49 = x

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Two angles of a triangle are 30°and 80°. Find the third angle.

stiv31 [10]

Answer:

The third angle of the triangle is 70°.

Step-by-step explanation:

The two angles of a triangle (say Δ ABC) are given to be 30° and 80°.

We have to find the third angle.

Let us assume that ∠ A = 30° and ∠ B = 80°, then we have to find ∠ C.

Now, we know that, ∠ A + ∠ B + ∠ C = 180° {Property of a triangle}

⇒ 30° + 80° + ∠ C = 180°

⇒ ∠ C = 180° - 30° - 80° = 70°

Therefore, the third angle of the triangle is 70°. (Answer)

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3 years ago

What is the value of 4.5 x 10^5 written in standard form?

soldier1979 [14.2K]

Answer:

4500000

Step-by-step explanation: Let me know if this helped

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3 years ago

Kyle and Taylor ran laps after school to train for the baseball team. The ratio of the number of laps Kyle ran to the number of

Zinaida [17]

Answer:

9 laps.

Step-by-step explanation:

-Let x be the number of laps Taylor runs.

-Given that the ratio of Kyle to Taylor laps is 2:3, we express the actual laps in meters to solve for x as follows:

$\frac{Kyle}{Taylor}=\frac{2}{3}\\\\\therefore \frac{2}{3}=\frac{6}{x}\\\\2x=18\\\\x=9$

Hence, Taylor runs 9 laps.

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3 years ago

PLEASE HELP!!!!!!!!!!!!!

Marianna [84]

Answer:

Step-by-step explanation:

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3 years ago

Give a 98% confidence interval for one population mean, given asample of 28 data points with sample mean 30.0 and sample standar

klio [65]

We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:

$\begin{gathered} 1-\alpha=0.98 \\ \alpha=0.02 \end{gathered}$

The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:

$CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack$

Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:

$CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack$

Where (from tables):

$Z_{0.99}=2.33$

Finally, the interval at 98% confidence level is:

$CI(\mu)=\lbrack28.94,31.06\rbrack$

4 0

1 year ago