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Kruka [31]
2 years ago
12

What is the y-intercept, axis of symmetry and vertex of the following function.

Mathematics
1 answer:
pshichka [43]2 years ago
6 0
The y intercept is always c (-5) and to the axis of symmetry u do -b/2a which is -1 and the vertex is (-1,10) because u pug in -1 into the equation. lmk if it’s right!
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How do you turn 245/360 into a decimal
erastovalidia [21]

Answer:

68.06%;

Step-by-step explanation:

6 0
2 years ago
Can i please get some help on this?
puteri [66]

Answer:

b

Step-by-step explanation:

The slope of b is 5.5 whereas the slope for the other three graphs are 5, so B is different from the rest.

A. (4, 0.8), so by rise over run, the slope is 4/0.8=5.

B. (55, 10), so 55/10=5.5.

C. (20, 4), so 20/4=5.

D. (50, 10), so 50/10=5.

Since B is the only graph with a slope of 5.5 and the other graphs have a slope of 5, the answer is B.

3 0
3 years ago
Triangle S R Q is shown. Angle S R Q is a right angle. An altitude is drawn from point R to point T on side S Q to form a right
juin [17]

Answer:

RT = 12 units

Step-by-step explanation:

From the figure attached,

ΔSRQ is right triangle.

m∠R = 90°

An altitude has been constructed from point T to side SQ.

m∠RTQ = 90°

By applying geometric mean theorem in triangle SRQ,

\frac{\text{RT}}{\text{ST}}=\frac{\text{TQ}}{\text{RT}}

\frac{x}{9}=\frac{16}{x}

x² = 16 × 9

x² = 144

x = √144

x = 12

Therefore, length of altitude RT is 12 units.

5 0
3 years ago
Combine like terms.
Elodia [21]
Answer is c: 24-3x 8+-11=-3
5 0
3 years ago
in a random sample of 200 people, 154 said that they watched educational television. fine the 90% confidence interval of the tru
melisa1 [442]

Answer:

[0.7210,0.8189] = [72.10%, 81.89%]

Step-by-step explanation:

The sample size is  

n = 200

the proportion is

p = 154/200 = 0.77

<em>Since both np ≥ 10 and n(1-p) ≥ 10 </em>

<em>We can approximate this discrete binomial distribution with the continuous Normal distribution. As the sample size is large enough, not applying the continuity correction factor makes no significant  difference. </em>

The approximation would be to a Normal curve with this parameters:

<em>Mean </em>

p = 0.77

<em>Standard deviation </em>

\bf s=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.77*0.23}{200}}=0.0298

The 90% confidence interval for the proportion would be then  

\bf  [0.77-z^*0.0298, 0.77+z^*0.0298]

where \bf z^* is the 10% critical value for the Normal N(0,1) distribution , this is a value such that the area under the N(0,1) curve outside the interval \bf [-z^*,z^*] is 10%=0.1

We can either use a table, a calculator or a spreadsheet to get this value.

In Excel or OpenOffice Calc we use the function

<em>NORMSINV(0.95) and we get a value of 1.645 </em>

The 90% confidence interval for the proportion is then

\bf  [0.77-1.645^*0.0298, 0.77+1.645^*0.0298]=[0.7210,0.8189]

This means there is a 90% probability that the proportion of people who watch educational television is between 72.10% and 81.89%

If the television company wanted to publicize the proportion of viewers, do you think it should use the 90% confidence interval?

Yes, I do.

5 0
3 years ago
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