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stepan [7]
3 years ago
6

HURRY PLEASE HELP ME I DONT KNOW WHAT TO DO

Mathematics
1 answer:
OleMash [197]3 years ago
5 0

Answer:

3rd option

Step-by-step explanation:

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Whould anyone mind helping please. The first box says 6.25x+12x=30.75 the second box says 6.25x+12=30.75 3rd box says 6.25x+30.7
Snowcat [4.5K]

Answer:

The correct equation is: 6x+12 = 30.75

And 3 people went to the movies

Step-by-step explanation:

We will write the equation to represent this scenario.

Let x be the number of people who went for movies.

Then the price of ticket for x people will be: 6.25x

They spent $12 on refreshments and a total of $30.75 so the equation will be:

6.25x+12=30.75

solving the equation will give us total number of people

So,

6.25x+12 = 30.75\\6.25x = 30.75-12\\6.25x = 18.75\\\frac{6.25x}{6.25} =\frac{18.75}{6.25}\\x = 3

Hence,

The correct equation is: 6x+12 = 30.75

And 3 people went to the movies

7 0
3 years ago
Convert –2y2 + x – 4y + 6 = 0 into standard form.
Rufina [12.5K]

Answer:A

Step-by-step explanation:

X=2(y+1)^2-6

4 0
3 years ago
Read 2 more answers
Which of the following radical expressions has an absolute value symbol in its simplified form?
umka21 [38]

Answer:

The answer is option b, its absolute value in simplified form gives

6\sqrt{2x^3} j

Step-by-step explanation:

Which of the following radical expressions has an absolute value symbol in its simplified form?

Answer option

6√64x^6

3VWjZbdGugPfWdqwB2+ZNNUbUB29afvQFuZASCHvF-8x^3

5√32x^5

4VWjZbdGugPfWdqwB2+ZNNUbUB29afvQFuZASCHvF16x

6√64x^6

to solve this

48x^3

b. 3√-8x^3

3*2√-2x^3

6√2x^3*√-1

6√2x^3j

which is a complex notation

5√32x^5

5*4√2x^5

20√2x^5

4√16x

4*4√x

16√x

6 0
4 years ago
The diameters of ball bearings are distributed normally. The mean diameter is 87 millimeters and the standard deviation is 6 mil
devlian [24]

Answer:

69.14% probability that the diameter of a selected bearing is greater than 84 millimeters

Step-by-step explanation:

According to the Question,

Given That, The diameters of ball bearings are distributed normally. The mean diameter is 87 millimeters and the standard deviation is 6 millimeters. Find the probability that the diameter of a selected bearing is greater than 84 millimeters.

  • In a set with mean and standard deviation, the Z score of a measure X is given by Z = (X-μ)/σ

we have μ=87 , σ=6 & X=84

  • Find the probability that the diameter of a selected bearing is greater than 84 millimeters

This is 1 subtracted by the p-value of Z when X = 84.

So, Z = (84-87)/6

Z = -3/6

Z = -0.5 has a p-value of 0.30854.

⇒1 - 0.30854 = 0.69146

  • 0.69146 = 69.14% probability that the diameter of a selected bearing is greater than 84 millimeters.

Note- (The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X)

7 0
3 years ago
Help please with algebra problem
Strike441 [17]
Number of weekend minutes used: x
Number of weekday minutes used: y

This month Nick was billed for 643 minutes:
(1) x+y=643

The charge for these minutes was $35.44
Telephone company charges $0.04 per minute for weekend calls (x)
and $0.08 per minute for calls made on weekdays (y)
(2) 0.04x+0.08y=35.44

We have a system of 2 equations and 2 unkowns:
(1) x+y=643
(2) 0.04x+0.08y=35.44

Using the method of substitution
Isolating x from the first equation:
(1) x+y-y=643-y
(3) x=643-y

Replacing x by 643-y in the second equation
(2) 0.04x+0.08y=35.44
0.04(643-y)+0.08y=35.44
25.72-0.04y+0.08y=35.44
0.04y+25.72=35.44

Solving for y:
0.04y+25.72-25.72=35.44-25.72
0.04y=9.72

Dividing both sides of the equation by 0.04:
0.04y/0.04=9.72/0.04
y=243

Replacing y by 243 in the equation (3)
(3) x=643-y
x=643-243
x=400


Answers:
The number of weekends minutes used was 400
The number of weekdays minutes used was 243
6 0
3 years ago
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