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3 years ago

0.5x10power5 equals

Mathematics

1 answer:

solniwko [45]3 years ago

4 0

Answer: 3125.

Step-by-step explanation: 0.5x10 is 5, then 5 to the fifth power is 3125 :)

Mark me as the brainliest if you think my answer was helpful!

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Please Help me ASAP i need to turn it in, in like 10 min. Please answer the questions in the screenshot ill give u 20 points Tha

Delvig [45]

Answer:

for #1 its the 1st and 4th box

for # 2 it's 13.5 and 20.25

Step-by-step explanation:

1st box - 10 ÷ 5 = 2 and 25 ÷ 5 = 5

4th box - 10 x 2 = 20 and 25 x 2 = 50

you have to make sure that you do the same thing for both sides of the equation

6 x 2.25 = 13.5

9 x 2.25 = 20.25

Dilation is an enlarging or shrinking of a math figure using a specific scale factor

multiply your numbers by that given scale factor to give you your answer of option 2

hope this is right I'm not completely positive about the send one.

4 0

3 years ago

If you cut 10 ounces of butter into 2/5 ounce pieces, how many

photoshop1234 [79]

2/5 plus 2/5 is five so we can assume if we multiply it by 4 it will be 10! 2/5 times 4 equals 10!!! Hope this helps! :D

6 0

2 years ago

How do you solve this

Orlov [11]

A) since the ratio of green peppers to pureed tomatoes is 1:4, the tomatoes have 4 times the amount of mL as the peppers, so there are 20mL of green peppers he should mix with 80 mL of pureed tomatoes.

8 0

3 years ago

Please Help<br><br> 2x^-3 = ?

Nesterboy [21]

2x^-3= 2/x^3
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4 0

3 years ago

Read 2 more answers

Silicon wafers are scored and then broken into the many small microchips that will he mounted into circuits. Two breaking method

antoniya [11.8K]

Answer:

The estimate is $- 0.0265\le K \le 0.0465$

Method B this is because the faulty breaks are less

Step-by-step explanation:

The number of microchips broken in method A is $n_1 = 400$

The number of faulty breaks of method A is $X_1 = 32$

The number of microchips broken in method B is $n_2 = 400$

The number of faulty breaks of method A is $X_2 = 32$

The proportion of the faulty breaks to the total breaks in method A is

$p_1 = \frac{32}{400}$

$p_1 = 0.08$

The proportion of the faulty to the total breaks in method B is

$p_2 = \frac{28}{400}$

$p_2 = 0.07$

For this estimation the standard error is

$SE = \sqrt{ \frac{p_1 (1 - p_1)}{n_1} + \frac{p_2 (1- p_2 )}{n_2} } }$

substituting values

$SE = \sqrt{ \frac{0.08 (1 - 0.08)}{400} + \frac{0.07 (1- 0.07 )}{400} } }$

$SE = 0.0186$

The z-values of confidence coefficient of 0.95 from the z-table is

$z_{0.95} = 1.96$

The difference between proportions of improperly broken microchips for the two breaking methods is mathematically represented as

$K = [p_1 - p_2 ] \pm z_{0.95} * SE$

substituting values

$K = [0.08 - 0.07 ] \pm 1.96 *0.0186$

$K = - 0.0265 \ or \ K = 0.0465$

The interval of the difference between proportions of improperly broken microchips for the two breaking methods is

$- 0.0265\le K \le 0.0465$

3 0

2 years ago