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goldenfox [79]
3 years ago
10

You can convert a temperature given in degrees Celsius to Fahrenheit temperature by using the expression 9x ÷ 5 + 32, where x is

the celsius temperature. Water freezes when the temperature is 0°C. At what fahrenheit temperature does water freeze?
Mathematics
1 answer:
Minchanka [31]3 years ago
4 0

Water freezes at 32 degree Fahrenheit.

Step-by-step explanation:

Given,

Formula for Celsius to Fahrenheit = \frac{9}{5}x+32

x = Celsius temperature

Boiling point of water = 0°C

We will put x=0

0°C = \frac{9}{5}(0)+32

0°C = 0+32

0°C = 32 degree Fahrenheit

Water freezes at 32 degree Fahrenheit.

Keywords: temperature, addition

Learn more about temperature at:

  • brainly.com/question/10894205
  • brainly.com/question/10987396

#LearnwithBrainly

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Scientific notation is written in the form a*10^n. What must be true about the a value?
Rama09 [41]

Answer:

It has to be a single digit between 1 and 10

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2 years ago
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Approximately how many degrees are in the measure of an interior angle of a regular eight sided polygon?
yan [13]
135° I think 1080( what the angles add up to) 1080÷8=135°
4 0
3 years ago
The diagram shows a straight line ABCD.
Nady [450]

Answer:

B (0, 285 ) , C (380, 0 )

Step-by-step explanation:

the first step is to obtain the equation of the line in slope- intercept form

y = mx + c ( m is the slope and c the y- intercept )

calculate m using the slope formula

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = A(- 260, 480 ) and (x₂, y₂ ) = D (620, - 180 )

m = \frac{-180-480}{620-(-260)} = \frac{-660}{620+260} = \frac{-660}{880} = - \frac{66}{88} = - 0.75 , then

y = - 0.75x + c ← is the partial equation

to find c substitute either of the 2 points into the partial equatio

using (620, - 180 ) , then

- 180 = - 465 + c ⇒ c = - 180 + 465 = 285

So y- intercept is B (0, 285 )

y = - 0.75x + 285 ← equation of line

to find the x- intercept , let y = 0 in the equation and solve for x

0 = - 0.75 + 285 ( subtract 285 from both sides )

- 285 = - 0.75x ( divide both sides by - 0.75 )

380 = x

x- intercept is C (380, 0 )

4 0
1 year ago
Read 2 more answers
Which equation can be used to find the value of b if side a measures 8.7 cm? 8.7 + b = 54.6 17.4 + b = 54.6 26.1 + b = 54.6 34.8
Gnom [1K]
The correct equation is 8.7 + b = 54.6
because it is given that measure of side a is 8.7cm, in all other equations the values of a is different.
from the first equation we can find the value of b, that is
b = 54.6 - 8.7 = 45.9
so, value of b is 45.9cm
5 0
2 years ago
. (0.5 point) We simulate the operations of a call center that opens from 8am to 6pm for 20 days. The daily average call waiting
SashulF [63]

Answer:

The 95% t-confidence interval for the difference in mean is approximately (-2.61, 1.16), therefore, there is not enough statistical evidence to show that there is a change in waiting time, therefore;

The change in the call waiting time is not statistically significant

Step-by-step explanation:

The given call waiting times are;

24.16, 20.17, 14.60, 19.79, 20.02, 14.60, 21.84, 21.45, 16.23, 19.60, 17.64, 16.53, 17.93, 22.81, 18.05, 16.36, 15.16, 19.24, 18.84, 20.77

19.81, 18.39, 24.34, 22.63, 20.20, 23.35, 16.21, 21.73, 17.18, 18.98, 19.35, 18.41, 20.57, 13.00, 17.25, 21.32, 23.29, 22.09, 12.88, 19.27

From the data we have;

The mean waiting time before the downsize, \overline x_1 = 18.7895

The mean waiting time before the downsize, s₁ = 2.705152

The sample size for the before the downsize, n₁ = 20

The mean waiting time after the downsize, \overline x_2 = 19.5125

The mean waiting time after the downsize, s₂ = 3.155945

The sample size for the after the downsize, n₂ = 20

The degrees of freedom, df = n₁ + n₂ - 2 = 20 + 20  - 2 = 38

df = 38

At 95% significance level, using a graphing calculator, we have; t_{\alpha /2} = ±2.026192

The t-confidence interval is given as follows;

\left (\bar{x}_{1}- \bar{x}_{2}  \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

Therefore;

\left (18.7895- 19.5152 \right )\pm 2.026192 \times \sqrt{\dfrac{2.705152^{2}}{20}+\dfrac{3.155945^2}{20}}

(18.7895 - 19.5125) - 2.026192*(2.705152²/20 + 3.155945²/20)^(0.5)

The 95% CI = -2.6063 < μ₂ - μ₁ < 1.16025996668

By approximation, we have;

The 95% CI = -2.61 < μ₂ - μ₁ < 1.16

Given that the 95% confidence interval ranges from a positive to a negative value, we are 95% sure that the confidence interval includes '0', therefore, there is sufficient evidence that there is no difference between the two means, and the change in call waiting time is not statistically significant.

6 0
2 years ago
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