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3 years ago

I need help with finding the arc measure of this problem.

Mathematics

1 answer:

Leto [7]3 years ago

4 0

Answer:

180+69=249

I got 180 from the ABC, that is half the circle then + the 69

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A business buys a computer for $3,000. After 4 years the value of the computer is expected to be $250. The value, v, can be rela

yuradex [85]

You are given two points in the linear function. At time 0 years, the value is $3000. At time 4 years, the value is $250. This means you have points (0, 3000) and (4, 250). You need to find the equation of the line that passes through those two points.

y = mx + b

m = (y2 - y1)/(x2 - x1) = (3000 - 250)/(0 - 4) = 2750/(-4) = -687.5

Use point (0, 3000).

3000 = -687.5(0) + b

b = 3000

The equation is

y = -687.5x + 3000

Since we are using points (t, v) instead of (x, y), we have:

v = -687.5t + 3000

Answer: d. v = -687.50 t + 3,000

6 0

3 years ago

A circus acrobat is shot out of a cannon. The equation for the acrobat's pathway can be modeled by h= -16t^2+25. How long will i

Alex_Xolod [135]

For this we almost have the following equation:
$h = -16t ^ 2 + 25
$
We must match the equation to zero:
$-16t ^ 2 + 25 = 0
$
From here, we clear the value of t.
We have then:
$16t ^ 2 = 25

t ^ 2 = 25/16$
We discard the negative root because we are looking for time:
$t = \sqrt{ \frac{25}{16} } 

$
$t = \frac{5}{4}$
Answer:
it will take him to reach the ground about:
$t = \frac{5}{4}$ seconds

5 0

3 years ago

I’ll give brainliest!! Help!!Jeremy deposits $24,000 into an account earning interest that is compounded monthly. The interest r

Schach [20]

The answer is $79,523.95

I just tooked the test.

4 0

3 years ago

Read 2 more answers

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

3x + 7 = -3x -2 How many solutions does this equation have

galina1969 [7]

This equation has one solution which is -1.5

3x+7=-3x-2
+2. +2
3x+9=-3x
-3x. -3x
9= -6x
/-6. /-6
X= -1.5

3 0

3 years ago