How many total outfit options are represented? 12 22 30 3

Choose all equations for which x=2 is a solution.

makvit [3.9K]

A and D that’s the answer

8 0

2 years ago

Someone help me with this

skad [1K]

Answer:

m∠ABC = 74

Step-by-step explanation:

The sum of the measures of three angles of any triangle is invariably equal to the straight angle, also expressed as 180 °

So you subtract the given measurements of ∠BAC (76°) and ∠ ACB (19°) from 180. 180-76-19 = 74

The measurement of angle ABC is 74°

7 0

3 years ago

(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

7 0

2 years ago

4-[2.5(5.4+2.6)-(-7.2)

valkas [14]

-23.2 I think this is the answer

7 0

3 years ago

A student is trying to solve the system of two equations given below: equation p: y z = 6 equation q: 3y 4z = 1 which of these i

Vadim26 [7]

Option C is correct (y + z = 6) ⋅ −3

What is a linear equation in math?

A linear equation only has one or two variables.
No variable in a linear equation is raised to a power greater than 1 or used as the denominator of a fraction.
When you find pairs of values that make a linear equation true and plot those pairs on a coordinate grid, all of the points lie on the same line.

As per the statement -

A student is trying to solve the system of two equations given below:

Equation P: y + z = 6 ....[1]

Equation Q: 3y + 4z = 1 ....[2]

Multiply the equation [1] by -3 to both sides we have;

-3 .( y + z = 6 ) ⇒ -3y -3z = -18..........(3)

Add equation [2] and [3] to eliminate the y-term;

z = -17

Therefore, the possible step used in eliminating the y-term is, (y + z = 6) ⋅ −3

Learn more about linear equation

brainly.com/question/11897796

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The complete question is -

A student is trying to solve the system of two equations given below: Equation P: y + z = 6 Equation Q: 3y + 4z = 1 Which of these is a possible step used in eliminating the y-term?

(y + z = 6) ⋅ 4

(3y + 4z = 1) ⋅ 4

(y + z = 6) ⋅ −3

(3y + 4z = 1) ⋅ 3

6 0

11 months ago

How many total outfit options are represented? 122230 3​

How many total outfit options are represented? 122230 3