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Ksenya-84 [330]
2 years ago
5

2(5x-1/3)=24x-7(2x+1) which statement about the type of solution is true?

Mathematics
1 answer:
TEA [102]2 years ago
4 0

Answer:

X = 19/3 (6.39)

Step-by-step explanation:

See the answer

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You obtain a dataset from a random sample. • You double-checked your dataset, and there were no typos, and no errors. • All cond
Pepsi [2]

Answer:

Interval is given with 97% confidence. Thus there is 3% probability that interval is not true.

Step-by-step explanation:

In Statistics, estimated intervals are given with some confidence level. In this example person develop the interval with 97% confidence.

<em>Statistically</em>, this means that the person can be 97% sure (not 100%)  that population mean is 74.3 < µ < 78.3. There is still 3% probability that population mean falls outside of the interval.

Small sample size may also lead wrong estimates.

7 0
3 years ago
0.02 as a mixed number in simplest form
BARSIC [14]
0.02 as a mixed number is 2/100 which is reduced to 1/50
6 0
3 years ago
On a trip to a lake, Kerrie and Shelly rode their bicycles four more than three times as many miles in the afternoon as in the m
Natalija [7]

Answer:

27 miles in the morning and 85 miles in the afternoon.

Explanation:

Let m be the number of miles they rode in the morning.  They rode 4 more than 3 times this many in the afternoon; this gives us the expression

3m+4

to represent the miles ridden in the afternoon.

Together they rode 112 miles; this means we add the morning miles, m, to the afternoon miles, 3m+4, and get 112:

m+3m+4 = 112

Combine like terms:

4m+4 = 112

Subtract 4 from each side:

4m+4-4= 112-4

4m = 108

Divide both sides by 4:

4m/4 = 108/4

m = 27

They rode 27 miles in the morning.

That means in the afternoon, they rode

3m+4 = 3(27)+4 = 81+4 = 85 miles.

5 0
3 years ago
Read 2 more answers
Can anyone help me with this please???
yanalaym [24]
Triangle = 3, Square = -1, Star = 5
8 0
2 years ago
Sample spaces For each of the following, list the sample space and tell whether you think the events are equally likely:
Schach [20]

Answer and explanation:

To find : List the sample space and tell whether you think the events are equally likely ?

Solution :

a) Toss 2 coins; record the order of heads and tails.

Let H is getting head and t is getting tail.

When two coins are tossed the sample space is {HH,HT,TH,TT}.

Total number of outcome = 4

As the outcome HT is different from TH. Each outcome is unique.

Events are equally likely since their probabilities \frac{1}{4} are same.

b) A family has 3 children; record the number of boys.

Let B denote boy and G denote girl.

If there are 3 children then the sample space is

{GGG,GGB,GBG,BGG,BBG,GBB,BGB,BBB}

The possible number of boys are 0,1,2 and 3.

Number of boys      Favorable outcome    Probability

           0                      GGG                        \frac{1}{8}

           1                    GGB,GBG,BGG          \frac{3}{8}

           2                   GBB,BGB,BBG           \frac{3}{8}

           3                       BBB                         \frac{1}{8}

Since the probabilities are not equal the events are not equally likely.

c)  Flip a coin until you get a head or 3 consecutive tails; record each flip.

Getting a head in a trial is dependent on the previous toss.

Similarly getting 3 consecutive tails also dependent on previous toss.

Hence, the probabilities cannot be equal and events cannot be equally likely.

d) Roll two dice; record the larger number

The sample space of rolling two dice is

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Now we form a table that the number of time each number occurs as maximum number then we find probability,

Highest number        Number of times         Probability

           1                                   1                     \frac{1}{36}

           2                                  3                    \frac{3}{36}

           3                                  5                    \frac{5}{36}

           4                                  7                    \frac{7}{36}

           5                                  9                    \frac{9}{36}

           6                                  11                    \frac{11}{36}

Since the probabilities are not the same the events are not equally likely.

4 0
3 years ago
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