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3 years ago

Deleted question.....

Mathematics

1 answer:

mamaluj [8]3 years ago

7 0

Answer:

i dont know what you're doing here?

Step-by-step explanation:

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Help I’ve been working on this problem for about 2 days and I still don’t get it and please explain and show me how to do the wo

Bond [772]

the boiling point of jet fuel is 329 degrees F. Rounded to the nearest degree, what is the temperature in degrees Celcius? Use the Formula: C = 5/9(F - 32), where C represents degrees celcius, and F represents degrees fahrenheit.

A). 165 degrees C

B). 183 degrees C

C). 201 degrees C

D). 535 degrees C

Solve:

C = 5/9(F - 32)

Simplify both sides:

C = 5/9(329 - 32) = 0

Simplify:

5/9

Equation at end of step 1:

C - (5/9 * 297) = 0

C - 165 = 0

Solve single variable:

C - 165 = 0

Add 165 to both sides:

C = 165

Therefore, Your answer would be, Letter Choice (A), (C = 165 degrees C

Hope that helps!!!!! Have a great day!!!!! : )

4 0

3 years ago

Solve the equation. 5x + 8 - 3x = -10 x = -1 x=1 x=9

aksik [14]

Answer:

solution,

 $5x + 8 - 3x = - 10 \\ or \: 5x - 3x + 8 = -10 \\ or \: 2x + 8 = -10 \\ or \: 2x = -10 - 8 \\ or \: 2x = -18\\ or \: x = \frac{-18}{2 } \\ x = -9$ 

hope this helps..

Good luck on your assignment

7 0

4 years ago

Read 2 more answers

Which of these is a number that can be expressed as a nonrepeating, nonterminating decimal?

faust18 [17]

Answer:

an irrational number

Step-by-step explanation:

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; (49) < μ₁ - μ₂ < (289)

Step-by-step explanation:

The given data are;

Bakery A

$\overline x_1$ = 1,880 cal

s₁ = 148 cal

n₁ = 10

Bakery B

$\overline x_2$ = 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;

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