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mote1985 [20]
3 years ago
12

Rewrite 64^18z^12 as a power of a product.

Mathematics
2 answers:
just olya [345]3 years ago
5 0
Yeah get Photomath to help you I don’t know if I spelled that right
tamaranim1 [39]3 years ago
4 0
Get photo math, easy questions like that will be easy
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A)<br>А<br>3.6 cm<br>3.2 cm<br>B В<br>С<br>4.5 cm​
uranmaximum [27]
What is the question?
6 0
3 years ago
The amount of toothpaste in a tube is normally distributed with a mean of 6.5 ounces and a standard
Anarel [89]

Answer:

a) 266 tubes ,  TC_r = $53.2

b) 266 tubes ,  T.Loss = $13.30

Step-by-step explanation:

Given:

- The sample size of tubes n = 1,000 tubes

- The mean of the sample u = 6.5 oz

- The standard deviation of the sample s.d = 0.8 oz

- Cost of manufacturing a tube C_t = 50 cents

- Cost of refilling a tube C_r = 20 cents

- Profit loss per tube Loss = 5 cents

Find:

a). How many tubes will be found to contain less than 6 ounces? In that case, what will be the total cost of the  refill?

b) How many tubes will be found to contain more than 7 ounces? In that case, what will be the amount of  profit lost?

Solution:

- First we will compute the probability of tube containing less than 6 oz.

- Declaring X : The amount of toothpaste.

Where,                         X ~ N ( 6.5 , 0.8 )

- We need to compute P ( X < 6 oz )?

Compute the Z-score value:

                  P ( X < 6 oz ) =  P ( Z < (6 - 6.5) / 0.8 ) = P ( Z < -0.625 )

Use the Z table to find the probability:

                               P ( X < 6 oz ) = P ( Z < -0.625 ) = 0.266

- The probability that it lies below 6 ounces. The total sample size is n = 1000.

       The number of tubes with X < 6 ounces = 1000* P ( X < 6 oz )

                                                                           = 1000*0.266 = 266 tubes.

- The total cost of refill:

                            TC_r = C_f*(number of tubes with X < 6)

                            TC_r = 20*266 = 5320 cents = $53.2

- We need to compute P ( X > 7 oz )?

Compute the Z-score value:

                  P ( X > 7 oz ) =  P ( Z > (7 - 6.5) / 0.8 ) = P ( Z < 0.625 )

Use the Z table to find the probability:

                               P ( X > 7 oz ) = P ( Z > 0.625 ) = 0.266

- The probability that it lies above 7 ounces. The total sample size is n = 1000.

       The number of tubes with X > 7 ounces = 1000* P ( X > 7 oz )

                                                                           = 1000*0.266 = 266 tubes.

- The total cost of refill:

                            T.Loss = Loss*(number of tubes with X > 7)

                            T.Loss = 5*266 = 1330 cents = $13.30

5 0
3 years ago
What’s the correct answer for this?
Burka [1]

Answer:

AP = 10

Step-by-step explanation:

According to secant-secant theorem

(PB)(AP)=(PD)(PC)

6(AP) = (5)(12)

AP = 60/6

AP = 10

8 0
3 years ago
Read 2 more answers
Robert and Danielle go to Macys in the mall. Robert buys 4 marroon polos and 3 khaki pants. Danielle buys 6 marroon polos and 4
Fofino [41]
C.
I took the test. Good luck
3 0
3 years ago
Need help simplifying this
diamong [38]

The simplified answer is \frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}.

<u>Step-by-step explanation:</u>

$\frac{3 y+2 x}{z+2 x}-\frac{2 y-3 x}{3 x+y}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}

To add or subtract denominators of the fraction must be same.

If it is not the same, we must take LCM of the denominators. and so we can add the fractions.

To make the denominator same multiply the 1st term (\frac{3x+y}{3x+y}) and 2nd term by (\frac{z+2x}{z+2x})

= \frac{(3 y+2 x)(3 x+y)}{(z+2 x)(3 x+y)}-\frac{(2 y-3 x)(z+2 x)}{(3 x+y)(z+2 x)}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}

LCM of the denominators is 6x²+ 3xz + 2xy +yz.

Multiply the factors in the numerator.

= \frac{\left(6 x^{2}+3 y^{2}+11 x y\right)}{(z+2 x)(3 x+y)}-\frac{\left(2 y z+4 x y-3 x z-6 x^{2}\right)}{(3 x+y)(z+2 x)}-\frac{2 z y+6 x z}{6 x^{2}+3 x z+2 x y+y z}

Now, the denominators are same, you can subtract it.

= \frac{\left(6 x^{2}+6 x^{2}+11 x y-4 x y-2 y z-2 y z+3 x z-6 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}

= \frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}

Thus the simplified solution is  \frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}

4 0
4 years ago
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