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3 years ago

7a/3a+3-5/4a-4=3a/2a+2

Mathematics

1 answer:

jekas [21]3 years ago

4 0

I guess i would be a=2.790220576310317 or −1.2902205763103165

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3x+2=11 plz hep im slow in math

stealth61 [152]

Answer:

Step-by-step explanation:

3×3=9+2=11. so 3 is the answer.

3 0

3 years ago

If vertex A is at (-1, 2) and vertex B is at (1, 5), then vertex A' is at and vertex B' is at .

lukranit [14]

The answer would be -0.5

5 0

3 years ago

laryy has 3 packets of ballons with 15 ballons in each packet. he wants to put the same number of balloons into 10 gift bags wha

wel

larry has 3 packer sin each 15 and total are 45 u wish put in 10 bags then divide 45 by 10 anser is 5

5 0

3 years ago

Use the net to compute the surface area of the three-dimensional figure.

Grace [21]

9514 1404 393

Answer:

382 square units

Step-by-step explanation:

The central four rectangles down the middle of the net are 9 units wide, and alternate between 8 and 7 units high. Then the area of those four rectangles is ...

9(8+7+8+7) = 270 . . . square units

The rectangles making up the two left and right "wings" of the net are 8 units high and 7 units wide, so have a total area of ...

2×(8)(7) = 112 . . . square units

Then the area of the figure computed from the net is ...

270 +112 = 382 . . . square units

Additional comment

You can reject the first two answer choices immediately, because they are odd. Each face will have an area that is the product of integers, so will be an integer. There are two faces of each size, so the total area of this figure must be an even number.

You may recognize that the dimensions are 8, 8+1, 8-1. Then the area is roughly that of a cube with dimensions of 8: 6×8² = 384. If you use these values (8, 8+1, 8-1) in the area formula, you find the area is actually 384-2 = 382. That area formula is A = 2(LW +H(L+W)).

3 0

3 years ago

From a large number of actuarial exam scores, a random sample of scores is selected, and it is found that of these are passing s

Mnenie [13.5K]

Supposing 60 out of 100 scores are passing scores, the 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

The lower limit is 0.5.
The upper limit is 0.7.

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

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

In which

z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

60 out of 100 scores are passing scores, hence $n = 100, \pi = \frac{60}{100} = 0.6$

95% confidence level

So $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so $z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 - 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.5$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 + 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.7$

The 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

The lower limit is 0.5.
The upper limit is 0.7.

A similar problem is given at brainly.com/question/16807970

5 0

2 years ago