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3 years ago

Find the value of a for which the solution of the inequality is all real numbers

Mathematics

1 answer:

Katen [24]3 years ago

8 0

Answer:

a= $\frac{4x+15}{x-3}$

Step-by-step explanation:

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Can someone pls help I can do this

GrogVix [38]

Answer:

these are supplementary

and x = 50

Step-by-step explanation:

one straight line = 180 degrees

there are only two angles making up this 180 degree line, so by definition they are supplementary

76 + 2x + 4 = 180

80 + 2x = 180

2x = 100

x = 50

4 0

3 years ago

The ratio of adults to students on the field trip is 3 to 5. If there are 15 students on the field trip, how many adults are the

Kazeer [188]

Answer:

There are 9 adults

Step-by-step explanation:

3:5 = x:15

3/5 = x/15

3(15) = 5x

5x = 45

x = 9 adults

3:5 = 9:15

4 0

3 years ago

dwight can drive 60 miles for each gallon of gas on his motorcycle how far can dwight go if he has 3 gallons of gas

Aleksandr-060686 [28]

Answer:

180

Step-by-step explanation: 60+60 =120 120+6

5 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

Does 3.9 x 6.12 equal 238.68.

STatiana [176]

23.868 but thats close one dp. off

3 0

3 years ago