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garik1379 [7]
3 years ago
11

Help please I have no clue how to do this please show work thanks

Mathematics
1 answer:
bazaltina [42]3 years ago
6 0

Terms that have the same variable part are called like terms. Like terms can be added or subtracted to form a single term.

1.

16x - 4x = -48

First, combine the like terms 16x and -4x.

12x = -48

Now divide both sides by 12.

x = -4

2.

7m - 5 - 13m = 25

First, combine the like terms 7m and -13m.

-6m - 5 = 25

Now add 5 to both sides.

-6m = 30

Divide both sides by -6.

m = -5

3.

12.25 = 0.5q + 3.75

Subtract 3.75 from both sides.

8.5 = 0.5q

Multiply both sides by 2.

17 = q

q = 17

4.

2(2x - 4) + x = 7

Distribute the 2.

4x - 8 + x = 7

Combine 4x and x.

5x - 8 = 7

Add 8 to both sides.

5x = 15

Divide both sides by 5.

x = 3

5.

8 = 3(3x + 8) - x

Distribute the 3.

8 = 9x + 24 - x

Combine 9x and -x.

8 = 8x + 24

Subtract 24 from both sides.

-16 = 8x

Divide both sides by 8.

-2 = x

x = -2

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A study9 compared personality characteristics between adult children of alcoholics and a control group matched on age and gender
34kurt

Answer:

df.= 28

P - value = 0.0124 < 0.05

There is a difference between the mean of  the well-being measure for the children of alcoholics, and a mean of the control group.

Step-by-step explanation:

As the given test is dependent or paired test the degrees of freedom is n-1 = 20-1 = 28

The reported t = 2.67

The P value for ∝= 0.05 at  calculated t= 2.67 and 28 d.f is 0.0124 which is less than 0.05 .

The null hypothesis is rejected.

The null hypothesis is that there is no difference between the means of the two samples.

H0: u1 = u2

The alternate hypotheis is accepted that there is  difference between the means of the two samples.

Ha: u1≠ u2

There is a difference between the mean of  the well-being measure for the children of alcoholics, and a mean of the control group.

4 0
3 years ago
A month of the year is chosen at random. What is the probability that it has 31 days
pogonyaev

There are 12 months, 7/12 have 31 days. You would have a 7/12 probability of choosing one with 31 days if it is random.

3 0
3 years ago
Read 2 more answers
Describe which of the two points lies on the graph of the line<br> Y-x=4<br> A. 9,5<br> B. 5,9
klasskru [66]
B. 5, 9

x = 5
y = 9

y - x = 4

(9) - (5) = 4

True

hope this helps
4 0
3 years ago
Ten percent of the engines manufactured on an assembly line are defective.
antoniya [11.8K]

Answer:

a) the probability that the first non-defective engine will be found on the second trial is 0.09

b) probability that the third non-defective engine will be found on the fifth trial is 0.00486

c) the Mean is 1.1111  and Variance is 0.1235  

d) the Mean is 3.3333and Variance is 0.3704  

Step-by-step explanation:

Firstly;

Let x be the number of trail on which the rth defective occurs

Also let the probability that an occurrence of a defective be p = 0.10

Here, x follows negative binomial distribution with parameters r and p

The probability mass function of X is as follows:

P(X = x) = [ x -1  p^r ( 1 - p)^(x-r)       ; x = r, r + 1, r + 2, ...

                 r - 1 ]

=   [ x -1  (0.10)^r ( 1 - 0.10)^(x-r)

     r - 1 ]

= [ x -1  (0.10)^r ( 0.9)^(x-r)  ..........n 1..... let this be equatio

    r - 1 ]

This represents the probability that the rth success occurs on the xth trail.  

a)

probability that the first non-defective engine will be found on the second trial?

Substitute r = 1 and x = 2 in equation 1  

P(X = 2)  = [ 2 - 1   (0.10)¹ ( 0.90 )²⁻¹

                   1 - 1 ]

= (0.10) × (0.90)

= 0.09

Therefore, the probability that the first non-defective engine will be found on the second trial is 0.09

b)

probability that the third non-defective engine will be found on the fifth trial?

So we substitute r = 3 and x = 5 in equation 1  

P(X = 5) = [ 5 - 1  (0.10)³ ( 0.90)⁵⁻³

                  3 - 1 ]

=    [ 4    (0.10)³ ( 0.9)²

      2 ]

= 0.00486

Therefore, probability that the third non-defective engine will be found on the fifth trial is 0.00486

Now the formula for the mean of the negative binomial distribution is as follows:  

Mean u = r / p    ------- let this be equation 2

The formula for the variance of the negative binomial distribution also is as follows:  

Variance α² = rq / p²   ---------- let this be equation 3

so

c)

the mean and variance of the number of trials on which the first non-defective engine is found.  

First, let the probability that non-defective engine found be p = 0.90

And q = (1 - p) = 1 - 0.90 = 0.10  

Now we substitute r = 1, p = 0.90 and q = 0.10 in equation 2 & 3simultaneously,

the mean and variances are as follows;

Mean = r/p = 1/0.90 = 1.1111

Variance = rq/p² = (1)(0.10) / (0.90)² = 0.1235  

Therefore the Mean is 1.1111  and Variance is 0.1235  

d)

the mean and variance of the number of trials on which the third non-defective engine is found  

Let the probability that non-defective engine found be p = 0.90

And q = (1 - p) = 1 - 0.90 = 0.10

Now we substitute r = 3, p = 0.90 and q = 0.10 in equation (2) and (3) simultaneously,

the mean and variances are;

Mean = r/p = 3/0.90 = 3.3333

Variance = rq/p² = (3)(0.10) / (0.90)² = 0.3704

Therefore the Mean is 3.3333 and Variance is 0.3704    

5 0
3 years ago
9^2= 9.5^2 + 4^2 - 2 (9.5) (4) (cosR)<br><br> I'm trying to find out the angle measure of R
alexandr402 [8]
Sorry I can't solve it.

Firstly, the equation is wrong. It is supposed to be <span>81 ≠ </span><span>30.25

Second of all, this is the best I got </span>81= <span><span>−<span>76 <span>(<span>cos<span>(r) </span></span></span></span></span>+ <span>106.25</span></span>
5 0
3 years ago
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