(help) what type of triangle is shown below

The y-intercept of g(x) in g(x) = 3x - 2 is (0,-2) (0,0) (2/3,0)

Dennis_Churaev [7]

For any y-intercept, the x-coordinate is zero. This is something to bear in mind to start with.

Now to find the y-intercept, find g(0) = -2.

Therefore the answer is (0,-2)

6 0

3 years ago

Read 2 more answers

Please help I’ll mark brainliest as long as you tell me how you got that answer, also I apologize if it’s kinda blurry. I hope y

kolezko [41]

Answer:

B.

Step-by-step explanation:

Not another one! ;)

A can be eliminated because the numbers don‘t match. B seems like a possible answer, let’s first check the others out. C and D can be slim instead because the variable, g, should be next to the 1.5 and 2.5. C and D both change the variable location. So, B.

4 0

2 years ago

how can u stand applying math to every thing doesn't it make every thing repetitive and why does every thing need a pattern does

Dominik [7]

Because math sucks like really really sucks

5 0

2 years ago

Look at the figure. Name the postulate or theorem you can use to prove the triangles congruent.

valentinak56 [21]

Answer:

AAS Theorem

Step-by-step explanation:

we can see that

angle ZVY ≅ angle WVY

angle VZY ≅ angle VWY

side VY shared by both triangles.

by Angle- Angle- SIde (AAS), the triangles are congruent.

4 0

3 years ago

Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r

ZanzabumX [31]

Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean weight = $\frac{\sum X}{n}$ = 21.88 ounces

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 0.201 ounces

n = sample of boxes = 12

$\mu$ = population mean weight

Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.796 < $t_1_1$ < 1.796) = 0.90 {As the critical value of t at 11 degrees of

freedom are -1.796 & 1.796 with P = 5%}

P(-1.796 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.796) = 0.90

P( $-1.796 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $21.88-1.796 \times {\frac{0.201}{\sqrt{12} } }$ , $21.88+1.796 \times {\frac{0.201}{\sqrt{12} } }$ ]

= [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

8 0

3 years ago