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3 years ago

What is the equation of the line that passes through the point (−1,−5) and has a slope of −3?

Mathematics

1 answer:

Hitman42 [59]3 years ago

6 0

Answer:

$y=-3x-8$

Step-by-step explanation:

Linear equations are typically organized in slope-intercept form:

$y=mx+b$ where $m$ is the slope and $b$ is the y-intercept (the value of y when the line crosses the y-axis)

<u>1) Plug the slope (m) into the equation</u>

We're given that the slope is -3. Plug -3 into the equation

$y=-3x+b$

<u>2) Solve for the y-intercept (b)</u>

To solve for b, plug the given point (-1,-5) into the equation as (x,y).

$-5=-3(-1)+b\\-5=3+b$

Subtract 3 from both sides

$-5-3=3+b-3\\-8=b$

Therefore, the y-intercept is -8. Plug -8 back into our original equation as b

$y=-3x-8$

I hope this helps!

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Approximate f by a Taylor polynomial with degree n at the number a. Step 1 The Taylor polynomial with degree n = 3 is T3(x) = f(

MAVERICK [17]

Answer:

If you center the series at x=1

$T_3(x) = e^2 + 4e^2 (x-1)+10(x-1)^2 + \frac{56}{3} e^2(x-1)^3 + R(x)$

Where $R(x)$ is the error.

Step-by-step explanation:

From the information given we know that

$f(x) = e^{2x^2}$

$f'(x) = 4x e^{2x^2}$ (This comes from the chain rule )

$f^{(2)}(x) = 4e^{2x^2} (4x^2+1)$ (This comes from the chain rule and the product rule)

$f^{(3)}(x) = 16xe^{2x^2}(4x^2 + 3)$ (This comes from the chain rule and the product rule)

If you center the series at x=1 then

$T_3(x) = e^2 + 4e^2 (x-1)+10(x-1)^2 + \frac{56}{3} e^2(x-1)^3 + R(x)$

Where $R(x)$ is the error.

3 0

3 years ago

A prticular type of tennis racket comes in a midsize versionand an oversize version. sixty percent of all customers at acertain

svetlana [45]

Answer:

a) P(x≥6)=0.633

b) P(4≤x≤8)=0.8989 (one standard deviation from the mean).

c) P(x≤7)=0.8328

Step-by-step explanation:

a) We can model this a binomial experiment. The probability of success p is the proportion of customers that prefer the oversize version (p=0.60).

The number of trials is n=10, as they select 10 randomly customers.

We have to calculate the probability that at least 6 out of 10 prefer the oversize version.

This can be calculated using the binomial expression:

$P(x\geq6)=\sum_{k=6}^{10}P(k)=P(6)+P(7)+P(8)+P(9)+P(10)\\\\\\P(x=6) = \binom{10}{6} p^{6}q^{4}=210*0.0467*0.0256=0.2508\\\\P(x=7) = \binom{10}{7} p^{7}q^{3}=120*0.028*0.064=0.215\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\P(x=9) = \binom{10}{9} p^{9}q^{1}=10*0.0101*0.4=0.0403\\\\P(x=10) = \binom{10}{10} p^{10}q^{0}=1*0.006*1=0.006\\\\\\P(x\geq6)=0.2508+0.215+0.1209+0.0403+0.006=0.633$

b) We first have to calculate the standard deviation from the mean of the binomial distribution. This is expressed as:

$\sigma=\sqrt{np(1-p)}=\sqrt{10*0.6*0.4}=\sqrt{2.4}=1.55$

The mean of this distribution is:

$\mu=np=10*0.6=6$

As this is a discrete distribution, we have to use integer values for the random variable. We will approximate both values for the bound of the interval.

$LL=\mu-\sigma=6-1.55=4.45\approx4\\\\UL=\mu+\sigma=6+1.55=7.55\approx8$

The probability of having between 4 and 8 customers choosing the oversize version is:

$P(4\leq x\leq 8)=\sum_{k=4}^8P(k)=P(4)+P(5)+P(6)+P(7)+P(8)\\\\\\P(x=4) = \binom{10}{4} p^{4}q^{6}=210*0.1296*0.0041=0.1115\\\\P(x=5) = \binom{10}{5} p^{5}q^{5}=252*0.0778*0.0102=0.2007\\\\P(x=6) = \binom{10}{6} p^{6}q^{4}=210*0.0467*0.0256=0.2508\\\\P(x=7) = \binom{10}{7} p^{7}q^{3}=120*0.028*0.064=0.215\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\\\P(4\leq x\leq 8)=0.1115+0.2007+0.2508+0.215+0.1209=0.8989$

c. The probability that all of the next ten customers who want this racket can get the version they want from current stock means that at most 7 customers pick the oversize version.

Then, we have to calculate P(x≤7). We will, for simplicity, calculate this probability substracting P(x>7) from 1.

$P(x\leq7)=1-\sum_{k=8}^{10}P(k)=1-(P(8)+P(9)+P(10))\\\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\P(x=9) = \binom{10}{9} p^{9}q^{1}=10*0.0101*0.4=0.0403\\\\P(x=10) = \binom{10}{10} p^{10}q^{0}=1*0.006*1=0.006\\\\\\P(x\leq 7)=1-(0.1209+0.0403+0.006)=1-0.1672=0.8328$