Y = mx + b
slope(m) = -1
(7,2)...x = 7 and y = 2
now we sub and find b, the y int
2 = -1(7) + b
2 = -7 + b
2 + 7 = b
9 = b

so ur equation of this line is : y = -1x + 9.....can also be written as :
y = -x + 9....or if u need it in standard form, it is : x + y = 9

8 0

3 years ago

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A standard American Eskimo dog has a mean weight of 30 pounds with a standard deviation of 2 pounds. Assuming the weights of sta

nalin [4]

Answer:

Option 2 - Approximately 24–36 pounds

Step-by-step explanation:

Given : A standard American Eskimo dog has a mean weight of 30 pounds with a standard deviation of 2 pounds. Assuming the weights of standard Eskimo dogs are normally distributed.

To find : What range of weights would 99.7% of the dogs have?

Solution :

The range of 99.7% will lie between the mean ± 3 standard deviations.

We have given,

Mean weight of Eskimo dogs is $\mu=30$

Standard deviation of Eskimo dogs is $\sigma=2$

The range of weights would 99.7% of the dogs have,

$R=\mu\pm3\sigma$

$R=30\pm3(2)$

$R=30\pm6$

$R=30+6,30-6$

$R=36,24$

Therefore, The range is approximately, 24 - 36 pounds.

So, Option 2 is correct.

5 0

3 years ago

Determine whether the improper integral converges or diverges, and find the value of each that converges.

Temka [501]

Answer:

It diverges.

Step-by-step explanation:

We are given the inetegral: $\int\limits^{\infty}_2 \frac{1}{x} (\ln x)^2 dx$

$\int\limits^{\infty}_2 \frac{1}{x} (\ln x)^2 dx=\int\limits^{\infty}_2 (\ln x)^2 d(\ln x)=\\\\=\lim_{t \to \infty} \int\limits^t_2 (\ln x)^2d(\ln x)=\lim_{t \to \infty} \frac{(\ln t)^3}{3} |^t_2=\infty-\frac{(\ln 2)^3}{3} =\infty$

So it is divergent.

7 0

3 years ago

We’re is the blue dot on the number line?

BartSMP [9]

Answer:

thank u !! -4.9

7 0

2 years ago

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