(a) Complete the table of values for the graph x + y = 6

Mathematics

1 answer:

djyliett [7]3 years ago

4 0

Answer:

(0,6)

(1,5)

(2,4)

(3,3)

(4,2)

Step-by-step explanation:

To solve this table first convert the equation to slope-intercept form (y=mx+b), to do this solve for y in the equation. The slope-intercept form is y=-x+6, where -1 is the slope and 6 is the y-intercept. Then from that, you can get your first point, because you know the y-intercept is 6, therefore when x equals 0 y must be 6. Next, because the slope is -1 that means the y-value decreases by 1 every time the x-value increases by 1. So to find the rest of the y-values simply subtract 1 from the previous y-value.

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A researcher is interested in the lengths of Salvelinus fontinalis (brook trout), which are known to be approximately Normally d

kompoz [17]

Answer:

c. 0.1151

Step-by-step explanation:

d) the probability that, on average, fish are larger than 86 centimeters in length.The area under part of a normal probability curve is directly proportional to probability and the value is calculated as

z = (x₁−x) /σ

where z = propability of normal curve

x₁ = variate mean = 86cm

x = mean of 80cm

σ = standard deviation = 5cm

applying the formula,

z= (86-80)/5

z = 6/5 =1.2

Using a table of partial areas beneath the standardized normal curve (see Table of normal curve, a z-value of 1.2 corresponds to an area of 0.3849 between the mean value. but, because the standard curve has 0.5, then will minus 0.3849 from 0.5= 0.5 - 0.3849 = 0.1151

Thus the probability of a fish are larger than 86 centimeters in length is 0.1151

5 0

4 years ago

onsider two populations for which μ1 = 38, σ1 = 2, μ2 = 23, and σ2 = 4. Suppose that two independent random samples of sizes n1

levacccp [35]

Answer:

Center = 15

Spread = 0.599

Shape = Normal

Step-by-step explanation:

The provided information is:

$\mu_1=38 \ \ \ \ \ \ \ \ \mu_2 = 23\\\sigma_1 =2 \ \ \ \ \ \ \ \ \ \sigma_2=4\\n_1=45 \ \ \ \ \ \ \ \ n_2 =59$

Thus the center(mean) of the distribution is:

$\begin{aligned}Mean &= \mu_1-\mu_2\\&=38-23\\&=15\end{aligned}$

The spread (standard deviation) of the distribution is:

$\begin{aligned}\textrm{Standard deviation}&=\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\\&=\sqrt{\frac{4}{45}+\frac{16}{59}}\\&=0.599\end{aligned}$