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3 years ago

Solve for x. -ax+4b>9

2 answers:

8 0

Let's solve for a.(−a)(x)+4b>9Step 1: Add -4b to both sides. −ax+4b+−4b>9+−4b −ax>−4b+9
Step 2: Divide both sides by -x. −ax−x>−4b+9−x a<4b−9x

Fed [463]3 years ago

4 0

-ax+4b>9
-ax>9-4b
-x>(9-4b)/a
x<-(9-4b)/a
x<(4b-9)/a

Answer: x<-(9-4b)/a or x<(4b-9)/a

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Please show work 7r + 21 = 49r

gulaghasi [49]

Answer:

1/2 = r

Step-by-step explanation:

7r + 21 = 49r

Subtract 7r from each side

7r-7r + 21 = 49r-7r

21 = 42r

Divide each side by 2

21/42 = 42r/42

1/2 = r

5 0

3 years ago

Read 2 more answers

Math worth 10 points thx

bija089 [108]

what thy answer child

6 0

3 years ago

Which expression shows how 6 • 45 can be rewritten using the distributive property?. A. 6 • 40 + 6. B. 6 • 4 + 6 • 5. C. 6 • 40

svlad2 [7]

6*45 = 6*(40 + 5) = 6*40 + 6*5

Option C.

5 0

3 years ago

Read 2 more answers

Several years ago, 50% of parents who had children in grades K-12 were satisfied with the quality of education the students r

galina1969 [7]

Answer:

The 95% confidence interval for the proportion of parents that are satisfied with their children's education is (0.4118, 0.4618). 0.5 is not part of the confidence interval, so this represents evidence that parents' attitudes toward the quality of education have changed.

Step-by-step explanation:

We have to see if 50% = 0.5 is part of the confidence interval.

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 1095, \pi = \frac{478}{1095} = 0.4365$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4365 - 1.96\sqrt{\frac{0.4365*0.5635}{1095}} = 0.4118$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4365 + 1.96\sqrt{\frac{0.4365*0.5635}{1095}} = 0.4612$

5 0

3 years ago

In a recent test, you scored 1.0 standard deviations above the average score attained, and the scores are normally distributed.

sesenic [268]

Answer:

97.75 %

Step-by-step explanation:

You score 1 standard deviation above the mean

a) we have from left tail up to the mean 50 % of all values

b) We have in the interval [ μ - σ ; μ + σ ] 95.5 % of all possible values, if we divide that 95.5 by 2 we get 47.75 %, the part of the values that correspond to the right, then in order to get the % of students scored lower we add

50 + 47.75 = 97.75 %

So you can say that the porcentage of students that scored lower than you is 97.75 %

3 0

3 years ago