Since there are no specific requirements given other than expression written using exponents, I am just going to give some examples.
1. 5x^2 - 3
2. 8^5 - 15x + 7
3. 15x^10 - 8x^7 + 16x^5 + 17^3 - 4^2 +13x - 2
Answer:
smaller=x
larger=x+26
total=90
x+x+26=90
2x=64
x=32
<h2>the smaller=32</h2><h2>the larger =58</h2>
Answer:
NO
Step-by-step explanation:
The changeability of a sampling distribution is measured by its variance or its standard deviation. The changeability of a sampling distribution depends on three factors:
- N: The number of observations in the population.
- n: The number of observations in the sample.
- The way that the random sample is chosen.
We know the following about the sampling distribution of the mean. The mean of the sampling distribution (μ_x) is equal to the mean of the population (μ). And the standard error of the sampling distribution (σ_x) is determined by the standard deviation of the population (σ), the population size (N), and the sample size (n). That is
μ_x=p
σ_x== [ σ / sqrt(n) ] * sqrt[ (N - n ) / (N - 1) ]
In the standard error formula, the factor sqrt[ (N - n ) / (N - 1) ] is called the finite population correction. When the population size is very large relative to the sample size, the finite population correction is approximately equal to one; and the standard error formula can be approximated by:
σ_x = σ / sqrt(n).
Answer:
A=(21/5,-22/5) B=(5,-6)
Step-by-step explanation:
Answer:
<h2><em><u>2</u></em></h2>
Step-by-step explanation:
<em><u>To</u></em><em><u> </u></em><em><u>find</u></em><em><u> </u></em><em><u>value</u></em><em><u>:</u></em>
2x + 3z
<em><u>Given</u></em><em><u> </u></em><em><u>values</u></em><em><u>:</u></em>
x = 4, y = 3 and z = -2
<em><u>Solution</u></em><em><u>:</u></em>
2x + 3z
<em>(</em><em>Putting</em><em> </em><em>the</em><em> </em><em>values</em><em> </em><em>of</em><em> </em><em>x</em><em> </em><em>=</em><em> </em><em>4</em><em> </em><em>and</em><em> </em><em>z</em><em> </em><em>=</em><em> </em><em>-2</em><em>)</em>
= 2(4) + 3(-2)
= 8 - 6
= <em><u>2 (Ans)</u></em>
