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4 years ago

» What is the product? 0.9 x 100 =

Mathematics

2 answers:

Zepler [3.9K]4 years ago

4 0

The product of this is 90

AURORKA [14]4 years ago

4 0

Answer:

0.9 × 100 = 90

Step-by-step explanation:

When you multiply a value by 10, you move the decimal one place to the right. When you multiply by 100, you move the decimal two places to the right. And so on. Therefore, 0.9 × 100 means taking 0.9 and moving the decimal two places to the right.

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What does this mean and how do u solve it?

DochEvi [55]

The current Brainliest answer seems to be answering the question "Every integer is a multiple of which number?" rather than the question presented here.

We say that one number is a multiple of a second number if we can get to the first one by counting by the second. For example, 18 is a multiple of 6 because we can reach it by counting by 6's (6, 12, 18). Note that, for any number we want to count by, we can always start our count at 0.

By 2's: 0, 2, 4, 6, 8

By 6's: 0, 6, 12, 18

By 7's: 0, 7, 14, 21

Because we can always "reach" 0 regardless of the integer we're counting by, we can say that 0 is a multiple of every integer.

More formally, we say that some number n is a multiple of an integer x if we can find another integer y so that x · y = n. By this definition, 18 would be a multiple of 6 because 6 · 3 = 18, and 3 is an integer. We can use the property that the product of any number and 0 is 0 to say that x · 0 = 0, where x can be any integer we want. Since 0 is also an integer, this means that, by definition, 0 is a multiple of every integer.

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4 years ago

The table shows the shoe sizes of women of different ages. A 2-column table with 4 rows. The first column is labeled age with en

elena55 [62]

Answer:

A.) weak positive correlation

The person above just wanted to steal points.

Step-by-step explanation:

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3 years ago

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3 boxes of baseball cards weigh 90 pounds. How many pounds do 4 boxes weigh

Anvisha [2.4K]

120 pounds
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3 years ago

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. Suppose that the signal x(t) = 2e −t/3 + e −t + 3e −t/2 is measured in the presence of noise distributed uniformly in the inte

galben [10]

Step-by-step explanation:

An exponentially decaying signal is of the form x(t) = Ce^(-αt) in terms of an initial value C and a decay rate α > 0. The signal equals a fraction 1/e of its initial value after the characteristic time scale t = 1/α.

Given

x(t) = 2e^(-t/3) + e^(-t) + 3e^(-t/2)

The decay rates are: 1/3, 1, and 1/2.

The slowest decay rate α is the minimum of {1/3, 1, 1/2} = 1/3.

The corresponding time scale

is only 2/3 times larger than the next faster decay rate 1/2, so the decay rates are NOT well separated. The slowest component is larger than 0.5 as long as t < (ln3)/0.5 ≈ 2.2.

The other two components add

up to more than the value of the slowest component. We conclude that the component with slowest decay rate dominates measurements on any interval even in the presence of noise.

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3 years ago

TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs via digital streamin

choli [55]

Answer:

a) The 99% confidence interval would be given (0.523;0.577).

We are 99% confident that this interval contains the true population proportion.

b) $n=\frac{0.55(1-0.55)}{(\frac{0.03}{2.58})^2}=1830.51$

And rounded up we have that n=1831

Step-by-step explanation:

Data given and notation

n=2341 represent the random sample taken

X represent the people that they have watched digitally streamed TV programming on some type of device

$\hat p=0.55$ estimated proportion of people that they have watched digitally streamed TV programming on some type of device

$\alpha=0.01$ represent the significance level

Confidence =0.99 or 99%

z would represent the statistic for the confidence interval

p= population proportion of people that they have watched digitally streamed TV programming on some type of device

The population proportion present the following distribution:

$p \sim N (p, \sqrt{\frac{p(1-p)}{n}}$

Part a) Confidence interval

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.