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marishachu [46]
3 years ago
11

Write the number that match the clues

Mathematics
1 answer:
Mumz [18]3 years ago
5 0
308 well that's all too that.
You might be interested in
6x+32=−2x
velikii [3]
Whatever you do to one side, you have to do to the other.
Try to get all the x values and regular values together.
To do that, just subtract both sides by 6x to cancel it out on one side and subtract it from the other side:

6x-6x+32=-2x-6x
=
32=-8x
now divide both sides by -8 to get the x by itself

32/-8=-8/-8x
= 32/-8=x
-4=x
so your answer is: x= -4
5 0
3 years ago
Determine whether the statement describes a descriptive or inferential statistic. A survey of 3459 people revealed that 52% work
andriy [413]

Answer:

Inferential statistics.

Step-by-step explanation:

We have been given a statement. We are asked to determine whether the given statement describes a descriptive or inferential statistic.

Statement: A survey of 3459 people revealed that 52% work a full-time job; therefore it can be assumed that 52% of the U.S. population works a full-time job.

We know that descriptive statistics represents the description about population using the data through numerical calculations, graphs or tables. While inferential statistics makes predictions about a population based on a sample of data taken from the population.

Since our statement gives the prediction about people, who work full-time in U.S. based on a survey of 3459 U.S. citizens, therefore, the given statement is inferential statistic.

7 0
3 years ago
Solve the system by elimination -3x+4y=16 3x+y=4
Illusion [34]
X's cancel 5y = 20 y= 4 X=0
7 0
3 years ago
The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78
____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

m-t_{a/2,n-1}\frac{s}{\sqrt{n} } ≤ μ ≤ m+t_{a/2,n-1}\frac{s}{\sqrt{n} }

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and t_{a/2,n-1} is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and t_{0.05,9} by 1.8331, we get that the 90% confidence interval for the mean speed is:

74.2-(1.8331)\frac{5.3083}{\sqrt{10} } ≤ μ ≤ 74.2+(1.8331)\frac{5.3083}{\sqrt{10} }

74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077

71.123 ≤ μ ≤ 77.277

8 0
3 years ago
Meredith invested $5,500 into an account that earned 5% interest per year. When she cashed out the investment to use it as a dow
shtirl [24]

Answer:

Step-by-step explanation:

Using I = PRT/100

where P = The principal amount invested = $5500

R = Rate of Interest =5%

T = Time (in years)

I = Interest = $7,739.05 - $5500 = $2239.05

Substituting the above values in the formula

I = PRT/100 becomes

2239.05 = 7,739.05 * 5 * t / 100 -------- Multiply 100 to both sides

2239.05 * 100 = 100 * 7739.05 * 5 * t / 100

223905 = 7739.05 * 5 * t

223905 = 38695.25t

Divide both sides by 38695.25t

223905/38695.25 = t

5.786369 = t

T = 5.79 (approximated)

8 0
3 years ago
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