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3 years ago

How do I solve that equation

Mathematics

2 answers:

Masteriza [31]3 years ago

6 0

I belive it would be A

rewona [7]3 years ago

3 0

Solving: (5)(3 - 1) + 4

Step One: Subtract 1 from 3 which is 2
5(3 − 1) + 4

Step Two, Multiply 5 by 2 which is 10=(5)(2) + 4

Step Three, Add 4 to 10 which is 14=10+4

=14

Answer:
(A)14

Hope this helps!

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Circle the greater fraction then right and saw the subtraction problem to find the difference of the fractions 9/10 11/12

Anon25 [30]

11/12 is the greater fraction.

2/60

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4 years ago

Your job is to paint lines in a long, straight parking lot. You determine the dividing lines need to be parallel. The length of

kozerog [31]

Answer:

a. 9 ft

b. 90 ° right angled

c. Right angle

d. 90°

e, Right angle

f. Angles on a straight line

g. 18 spots

Step-by-step explanation:

Here we have maximization question;

a. The separation distance of the dividing lines in a parking lot need to be far apart enough as to accommodate a vehicle with room to open the doors, therefore, it should be between 8.5 to 10 ft wide which gives a mean parking space width of approximately 9 ft

b. The angle of lines of the parking lot to the curb that will accommodate the most cars is 90°, because it reduces the width occupied by a car

c. The angle is right angled

d. Since the adjacent angle + calculated angle = angles on a straight line = 180 °

Therefore, adjacent angle = 90°

e. The angle is right angled

f. Angles on a straight line

g. The number of spots will be 162/9 = 18 spots.

8 0

4 years ago

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cupoosta [38]

Answer:

3 is your answer im pretty sure:)))

Step-by-step explanation:

7 0

3 years ago

The second sample was the same size as the first, and the proportion of sales identified as organic was 0.4. How does the 95 per

enot [183]

Answer:

The summer interval is wider and has a greater point of estimate

Step-by-step explanation:

Assuming this problem: "An environmental group wanted to estimate the proportion of fresh produce sales indentified as organic in a grocery store. In the winter, the group obtained a random sample of sales from the store and used the data to construct a 95percent z-interval for a proportion (0.087,0.133). Six months later in the summer, the group obtained a second random sample of sales from the store. The second sample was the same size as the first, and the proportion of sales identified as organic was 0.4. How does the 95 percent z-interval for a proportion constructed from the summer sample compare to the winter interval? "

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

We can find the margin of error from the interval given on this case would be:

$ME= \frac{0.133-0.087}{2}=0.023$

And the point of estimate for the proportion obtained would be:

$\hat p = 0.133-0.023=0.11$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =0.023$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

$n=\frac{0.11(1-0.11)}{(\frac{0.023}{1.96})^2}=710.95$

And rounded up we have that n=711

Now we know that we use the same sample size for the new confidence interval. And replacing into the confidence interval formula we got:

$0.4 - 1.96 \sqrt{\frac{0.4(1-0.4)}{711}}=0.364$

$0.4 + 1.96 \sqrt{\frac{0.4(1-0.4)}{711}}=0.436$

And the new 95% confidence interval would be given (0.364;0.436) has a width of 0.436-0.364=0.072. And the width for the previous confidence interval is 0.133-0.087= 0.046. So then the best answer is:

The summer interval is wider and has a greater point of estimate

7 0

3 years ago

9 x 100 + 1 x 1/10 + 3 x 1/100

Bess [88]

Answer: 900.13, or 900 13/100

Step-by-step explanation: PEMDAS, multiply first, than add.

5 0

3 years ago

Read 2 more answers