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3 years ago

15

Ursula wrote the sum 5.815 plus 6.021 as a sum of two mixed numbers.

1 answer:

aleksandrvk [35]3 years ago

5 0

Sum of 5.815 + 6.021 = Mixed Numbers<span>Let’s convert decimal numbers to mixed numbers
=>5.815 the decimal is .815 and 6.021 the decimal is .021
=> .815/1000
=> .021/1000
So the sum she wrote is:
=> 5 815/1000 + 6 021/1000 = 11 836/1000 in mixed numbers

Compare the sum of Mix number to the sum of the decimals
Mixed numbers Sum = 11 836/1000
Decimal numbers Sum = 11.836 </span>

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Please Help! Are F(x) and G(x) inverse functions across the domain [3,+∞)?

allochka39001 [22]

Answer:

A

Explanation:

For functions to be inverse, it must be true that f( g(x) ) = x and g( f(x) ) = x.

But for F( G(x) ), we have √( G(x) - 3 ) + 8

= √( (x+8)² - 3 - 3 ) + 8

= √( (x+8)² - 6 ) + 8

This -6 part should be cancelled out for functions to work out but we cannot do that, therefore F(x) and G(x) are not inverse.

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3 years ago

(4.70x10^3)+(8.6x10^5)

Natali [406]

Answer:

8.647 x 10^5

Step-by-step explanation:

3 0

3 years ago

Estimate the sum 7.36 + 9.47

arsen [322]

Answer:

16.83

Step-by-step explanation:

7.36

+ 9.47

----------

16.83

4 0

3 years ago

How do i turn this to decimals?<br><br> 818/1320<br><br> 1742/2540<br><br> 396/1220

trapecia [35]

You simply divide numerator by denominator
818/1320 = 0.619
1742/2540 = 0.685
396/1220 = 0.32
hope this helps ;3

6 0

3 years ago

Read 2 more answers

When the sample size and the sample proportion remain the same, a 90 percent confidence interval for a population proportion p w

deff fn [24]

Answer:

A 90% confidence interval for <em>p</em> will be <u>narrower </u>than the 99% confidence interval.

Step-by-step explanation:

The formula to compute the (1 - <em>α</em>) % confidence interval for a population proportion is:

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

Here $\hat p$ is the sample proportion.

The margin of error of the confidence interval is:

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

The MOE is dependent on:

Confidence level
Standard deviation
Sample size

The MOE is directly related to the confidence level and standard deviation.

So if any of the two increases then the MOE also increases, thus widening the confidence interval.

And the MOE is inversely related to the sample size.

So if the sample increases the MOE decreases and vice versa.

It is provided that the sample size and the sample proportion are not altered.

The critical value of <em>z</em> for 90% confidence level is:

$z_{\alpha/2}= z_{0.10/2}=z_{0.05}=1.645$

And the critical value of <em>z</em> for 99% confidence level is:

$z_{\alpha/2}= z_{0.05/2}=z_{0.05}=1.96$

So as the confidence level increases the critical value increases.

Thus, a 90% confidence interval for <em>p</em> will be narrower than the 99% confidence interval.

6 0

2 years ago