All categories

3 years ago

Please help me out! :) !!!!!!

Mathematics

1 answer:

rjkz [21]3 years ago

4 0

Answer:

your answer is 8/40 but is reduced to 1/5

Step-by-step explanation:

You might be interested in

Please help, i really suck at geometry

Alex Ar [27]

Have you found the answer?

7 0

3 years ago

Read 2 more answers

. Only 2% of a large population of 100-ohm gold-band resistors have resistances that exceed 105 ohms. a. For samples of size 100

Soloha48 [4]

Using the information given above, the sampling distribution of the sample proportion of 100-ohm gold-band is 2.

Sampling distribution of proportion, P = 2% = 0.02

Sample size, n = 100

The sampling distribution of the sample proportion can be calculated thus:

Distribution of sample proportion = np

Distribution of sample proportion = (100 × 0.02) = 2

Therefore, there is a probability that only 2 of the samples will have resistances exceeding 105 ohms.

Learn more : brainly.com/question/18405415

4 0

2 years ago

2. He lives in a big city. a. b.

Verizon [17]

?? I don’t really get it, what should I do.

8 0

2 years ago

Help ASAP show work please thanksss!!!!

Llana [10]

Answer:

$\displaystyle log_\frac{1}{2}(64)=-6$

Step-by-step explanation:

Properties of Logarithms

We'll recall below the basic properties of logarithms:

$log_b(1) = 0$

Logarithm of the base:

$log_b(b) = 1$

Product rule:

$log_b(xy) = log_b(x) + log_b(y)$

Division rule:

$\displaystyle log_b(\frac{x}{y}) = log_b(x) - log_b(y)$

Power rule:

$log_b(x^n) = n\cdot log_b(x)$

Change of base:

$\displaystyle log_b(x) = \frac{ log_a(x)}{log_a(b)}$

Simplifying logarithms often requires the application of one or more of the above properties.

Simplify

$\displaystyle log_\frac{1}{2}(64)$

Factoring $64=2^6$ .

$\displaystyle log_\frac{1}{2}(64)=\displaystyle log_\frac{1}{2}(2^6)$

Applying the power rule:

$\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}(2)$

Since

$\displaystyle 2=(1/2)^{-1}$

$\displaystyle log_\frac{1}{2}(64)=6\cdot log_\frac{1}{2}((1/2)^{-1})$

Applying the power rule:

$\displaystyle log_\frac{1}{2}(64)=-6\cdot log_\frac{1}{2}(\frac{1}{2})$

Applying the logarithm of the base:

$\mathbf{\displaystyle log_\frac{1}{2}(64)=-6}$

5 0

2 years ago

How many centimeters are there in 1/5 of an inch

Lynna [10]

Answer:

0.508

Step-by-step explanation:

4 0

3 years ago