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3 years ago

How can you add, subtract, and multiply with decimal

1 answer:

5 0

To add or subtract you do what you would normally do then put
the decimal in the answer

to multiply you multiply the numbers then add the demical places from both
numbers then put the decimal there

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Linh has obtained scores of 82, 79, and 75 on her midterm algebra exams. If the final exam counts twice as much as a midterm, wh

OLga [1]

Answer:

She must make a score of 42 in the final exam to obtain an average of 80.

Step-by-step explanation:

Hi there!

Let x be the score of the final exam, the equation will be as follows (remember that the score of the final counts double):

(82 + 79 + 75 + 2x)/4 = 80

Let´s solve the equation for x multiplying each side of the equation be 4:

82 + 79 +75 + 2x = 320

Substract 82, 79 and 75 to both sides of the equation:

2x = 320 - 82 - 79 -75

2x = 84

x = 84/2

x = 42

She must make a score of 42 in the final exam to obtain an average of 80.

Have a nice day!

8 0

3 years ago

Solve this problem<br> PLS SOLVE ASAP!!

guajiro [1.7K]

α $a^{2} (-4+6)$

8 0

3 years ago

Please help me i really need it

hram777 [196]

Answer:

You just gotta multiply then divide it

Step-by-step explanation:

5 0

2 years ago

P, Q, V, and K are collinear with V between K and P, and Q between V and K. If VP = 14x + 4, PK = x + 630, VQ = 17x + 6, and KQ

Ierofanga [76]

It's given in the question,

P, Q, V and K are collinear.

VP = 14x + 4

PK = x + 630

VQ = 17x + 6

KQ = 11x + 5

By segment addition postulate,

KQ + VQ + VP = KP

Substitute the values in the expression,

(11x + 5) + (17x + 6) + (14x + 4) = x + 630

(11x + 17x + 14x) + (5 + 6 + 4) = x + 630

42x + 15 = x + 630

42x - x = 630 - 15

41x = 615

x = $\frac{615}{41}$

x = 15

Therefore, value of VP = (14x + 4)

= 14(15) + 4

= 214 units

Learn more,

brainly.com/question/628239

3 0

2 years ago

You recently sent out a survey to determine if the percentage of adults who use social media has changed from 66%, which was the

il63 [147K]

Answer:

The 98% confidence interval estimate of the proportion of adults who use social media is (0.56, 0.6034).

Step-by-step explanation:

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 $1 - \frac{\alpha}{2}$ .

Of the 2809 people who responded to survey, 1634 stated that they currently use social media.

This means that $n = 2809, \pi = \frac{1634}{2809} = 0.5817$

98% confidence level

So $\alpha = 0.02$ , z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$ , so $Z = 2.327$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5817 - 2.327\sqrt{\frac{0.5817*4183}{2809}} = 0.56$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5817 + 2.327\sqrt{\frac{0.5817*4183}{2809}} = 0.6034$

The 98% confidence interval estimate of the proportion of adults who use social media is (0.56, 0.6034).

6 0

2 years ago