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3 years ago

Aiden calculated the area of Triangle QRS finding half the product of 1.5 and 1.1. What error did Aiden make?

Mathematics

1 answer:

Gnom [1K]3 years ago

6 0

Aiden should have calculated
area = (1.1 * 5.4) / 2
because triangle area = half the base times the altitude.

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Write the standard equation of the circle with the given Center and radius:

Kipish [7]

9514 1404 393

Answer:

B. (x +4)² +(y +5)² = 64

Step-by-step explanation:

The equation for a circle with center (h, k) and radius r is ...

(x -h)² +(y -k)² = r²

Filling in your given values, you have ...

(x -(-4))² +(y -(-5))² = 8²

Simplifying, you get ...

(x +4)² +(y +5)² = 64 . . . . . . . matches choice B

_____

Additional comment

As you learn more different "standard form" equations, it may be helpful to keep a list, along with reminders of what the variables stand for and how the equation is used.

5 0

2 years ago

What is the number of acute obtuse and right angle of a rectangle

matrenka [14]

The number of a acute angle in a rectangle is 89 through 1 the obtuse angle in a rectangle is 91 through inf and a right angle of a rectangle is 90 degrees

4 0

3 years ago

PLS HELP ASAP THANKS ILL GIVE BRAINLKEST

Elis [28]

Answer:

Step-by-step explanation:

move constant

x+7 ≥ 12

subtract

x≥ 12-7

solution

x ≥ 5

4 0

2 years ago

Let us suppose we have data on the absorbency of paper towels that were produced by two different manufacturing processes. From

maksim [4K]

Answer:

The 95% CI for the difference of means is:

$-155.45 \leq \mu_1-\mu_2 \leq -44.55$

Step-by-step explanation:

The question is incomplete:

"Find a 95% confidence interval on the difference of the towels mean absorbency produced by the two processes. Assumed that the standard deviations are estimated from the data. Round to two decimals places."

Process 1:

- Sample size: 10

- Mean: 200

- S.D.: 15

Process 2:

- Sample size: 4

- Mean: 300

- S.D.: 50

The difference of the sample means is:

$M_d=M_1-M_2=200-300=-100$

The standard deviation can be estimated as:

$\sigma_d=\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\\\\\sigma_d=\sqrt{\frac{15^2}{10}+\frac{50^2}{4}} =\sqrt{22.5+625}=\sqrt{647.5}=25.45$

The degrees of freedom are:

$df=n_1+n_2-2=10+4-2=12$

The t-value for a 95% confidence interval and 12 degrees of freedom is t=±2.179.

Then, the confidence interval can be written as:

$M_d-t\cdot \sigma_d\leq \mu_1-\mu_2 \leq M_d+t\cdot \sigma_d\\\\-100-2.179*25.45\leq \mu_1-\mu_2 \leq -100+2.179*25.45\\\\-100-55.45 \leq \mu_1-\mu_2 \leq -100+55.45\\\\ -155.45 \leq \mu_1-\mu_2 \leq -44.55$

8 0

3 years ago

Is 1/2x-4 a function?

lakkis [162]

Answer:

Yes it is

Step-by-step explanation:

Each x value is represented by exactly one y value.

7 0

3 years ago