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andrey2020 [161]
3 years ago
12

Solve and find the side please

Mathematics
1 answer:
oee [108]3 years ago
3 0

Answer:

\displaystyle 1

Step-by-step explanation:

\displaystyle \frac{x}{0,6} = csc\:37 \hookrightarrow 0,6csc\:37 = x \hookrightarrow 0,9969840846... = x \\ \\ 1 ≈ x

<em>OR</em>

\displaystyle \frac{0,6}{x} = sin\:37 \hookrightarrow xsin\:37 = 0,6 \hookrightarrow \frac{0,6}{sin\:37} = x \hookrightarrow 0,9969840846... = x \\ \\ 1 ≈ x

Information on trigonometric ratios

\displaystyle \frac{OPPOCITE}{HYPOTENUSE} = sin\:θ \\ \frac{ADJACENT}{HYPOTENUSE} = cos\:θ \\ \frac{OPPOCITE}{ADJACENT} = tan\:θ \\ \frac{HYPOTENUSE}{ADJACENT} = sec\:θ \\ \frac{HYPOTENUSE}{OPPOCITE} = csc\:θ \\ \frac{ADJACENT}{OPPOCITE} = cot\:θ

I am joyous to assist you at any time.

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Answer:

12.56

Step-by-step explanation:

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2 years ago
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A manufacturer is interested in the output voltage of a power supply used in a PC. Output voltage is assumed to be normally dist
Luda [366]

The missing values in the question are shown in bold forms below.

A manufacturer is interested in the output voltage of a power supply used in a PC. Output voltage is assumed to be normally distributed, with standard deviation of 0.25 V, and the manufacturer wished to test \mathbf{ H_o : \mu = 10 V  \ against \ H_1 : \mu  \neq 10V}, using  n = 10 units. Statistical Tables and Charts

(a) The critical region is \mathbf{\overline X < 9.83} or \mathbf{\overline X < 10.17} . Find the value of  \mathbf{\alpha }

Answer:

∝ = 0.032   (to 3 decimal place)

Step-by-step explanation:

From the given information:

P(\overline X < 9.83 ) + P( X> 10.17) = \bigg (1- P \bigg ( \dfrac{X - \mu}{\dfrac{\sigma}{\sqrt{n}}} \bigg )< Z<  \bigg ( \dfrac{X - \mu}{\dfrac{\sigma}{\sqrt{n}}} \bigg ) \bigg )

P(\overline X < 9.83 ) + P( X> 10.17) = \bigg (1- P \bigg ( \dfrac{9.83 - 10}{\dfrac{0.25}{\sqrt{10}}} \bigg )< Z<  \bigg ( \dfrac{10.17- 10}{\dfrac{0.25}{\sqrt{10}}} \bigg ) \bigg )

P(\overline X < 9.83 ) + P( X> 10.17) = \bigg (1- P \bigg ( \dfrac{-0.17}{\dfrac{0.25}{\sqrt{10}}} \bigg )< Z<  \bigg ( \dfrac{0.17}{\dfrac{0.25}{\sqrt{10}}} \bigg ) \bigg )

P(\overline X < 9.83 ) + P( X> 10.17) = \bigg (1- P \bigg (-2.15 \bigg )< Z<  \bigg ( 2.15 \bigg ) \bigg )

From the z - tables;

P(\overline X < 9.83 ) + P( X> 10.17) = \bigg (1- (0.9842 -0.0158) \bigg )

\alpha = \mathbf{P(\overline X < 9.83 ) + P( X> 10.17) = 0.032}

5 0
3 years ago
A hotel pays the phone company $200 per month plus $0.35 for each call made. During January 6,000 calls were made. In February 4
Hunter-Best [27]

a. Calculate the hotel's phone bills for January and February.

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Total Cost = Fixed Cost + Variable Cost

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Total Cost = $200 + $0.30 x 10,000

Total Cost = $200 + $3,000

Total Cost = $3,200

February

Total Cost = Fixed Cost + Variable Cost

Total Cost = $200 + $0.30 x calls

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b. Calculate the cost per phone call in January and in February.

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Cost per Call = Total Cost / Total Calls

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c. Separate the January phone bill into its fixed and variable components.

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The fixed component is equal to $200 regardless of the total number of calls. The variable cost, therefore, is equal to the remaining $3,000 ($3,200 total - $200 fixed).

February

The fixed component is equal to $200 regardless of the total number of calls. The variable cost, therefore, is equal to the remaining $2,400 ($2,600 total - $200 fixed).

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1 year ago
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timurjin [86]

Answer:

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Step-by-step explanation:

\frac{13}{6} to mix fraction is  2 \frac{1}{6}

Hope this helps! :)

please mark this as brainiest

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I = Prt
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The interest rate is about 3.9%.
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