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Licemer1 [7]
3 years ago
5

X + 5y = 8 -x + 2y = -1 How to solve this?

Mathematics
1 answer:
ira [324]3 years ago
4 0

Answer:

x+5y=8

-5 -5

----------

x=3

-x+2y=-1

-2 -2

------------

-x

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4/3 x+12=-4 hell please right answer get brainliest points and show work please
IrinaVladis [17]

Answer:

x = -12

Step-by-step explanation:

4/3x + 12 = - 4   (multiply 3 both side)

4x + 36 = - 12 (minus 36 both side)

4x + 0 = -12

4x = -12 - 36

4x = - 48 (divide both side)

4x ÷ 4 = x

-48 ÷ 4 = -12

Answer: x = -12

8 0
3 years ago
How do I solve number 10? I'm not sure if I did it right.
grandymaker [24]
No, you did it right
3 0
2 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

<em>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.</em>

<u>So, 90% confidence interval for the population mean, </u>\mu<u> is ;</u>

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

<u>90% confidence interval for</u> \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
How do u solve to get x?
Vedmedyk [2.9K]

Step-by-step explanation:

2^(2x) - 5(2^x) = -4

(2^x)^2 - 5(2^x) = -4

Substitute 2^x = z --> x = ln(z)/ln(2)

z^2 - 5z = -4

z^2 - 5z + 4 = 0

(z - 1)(z - 4) = 0

z = 1 --> x = ln(1)/ln(2) = 0

z = 4 --> x = ln(4)/ln(2) = 2


4 0
3 years ago
You post a survey to the school about favorite desserts. Of the 540 students at the school, 60 answered the survey. 15 students
makkiz [27]

Answer:

135

Step-by-step explanation:

1. 540/60=9

2.9×15=135

4 0
3 years ago
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