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3 years ago

If (1,4) and (10,-68) are two anchor points on the trend line, then find the equation of the line

Mathematics

1 answer:

victus00 [196]3 years ago

3 0

Slope = -8

We plug it in :

4 = 1(-8) +b

B = 12

Y= -8x + 12

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write in slope intercept form an equation of the line that passes through the given points (-2,3) (2,7)

yarga [219]

1. Slope = Yb - Ya / Xb - Xa
= 7-3/2-(-2)
= 7-3/2+2
= 4/4
= 1

2. Y-intercept
Y = mx + b
7 = (1)(2) + b
7 = 2 + b
7 - 2 = 2-2 + b (two’s cancel out and we are just left with our b on the right side)
5 = B

Therefore, the equation is: Y = 1x + 5 which is also the same thing as: Y= X +5.

3 0

3 years ago

Difference

Ivan

Answer:

The value of a function is the actual calculation done at a certain point. The limit is - roughly speaking - the value at points that are “arbitrarily close” to the same point. For most commonly used functions, the value of a function at a point, and the limit at the same point, is the same - at least for most values.

4 0

3 years ago

Simplify 2x2a^2 x2a^2

horrorfan [7]

Step-by-step explanation:

> 2×2a²×2a²

8a⁴

> 36a³×1/4a²

9a³×1/a²

> 2⁶

> 5²m²

6 0

3 years ago

2 milliliters in one week

kipiarov [429]

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8 0

3 years ago

A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 421 gram setting. It is

Anastaziya [24]

Answer:

The p-value is approximately 0.0508

Step-by-step explanation:

The given parameters are;

The expected mass of banana chips bag filled by the machine, μ = 421 gams

The number of bags in the sample of bags filled by the machine, n = 26 bags

The mean mass of the bags in the sample, $\overline x$ = 413 grams

The standard deviation of the mass in the bags, s = 24 grams

The level of significance, α = 0.02

The null hypothesis, H₀; μ = 421 grams

The alternative hypothesis, Hₐ; μ < 421 grams

The test statistic is given as follows;

$t=\dfrac{\bar{x}-\mu }{\dfrac{s}{\sqrt{n}}}$

We get;

$t=\dfrac{413-421 }{\dfrac{24}{\sqrt{26}}} \approx -1.6996731712$

The t-value = -1.6996731712

The degrees of freedom df = n - 1 = 26 - 1 = 25

From an online source, we get the critical-t = 2.166587

The p-value is obtained using an online source as p(t ≤ -1.6996731712) = 0.0508

From the p-value table, we get

0.05 < p < 0.1

3 0

3 years ago