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3 years ago

(26 points)

Mathematics

2 answers:

Fed [463]3 years ago

8 0

Answer:

Given 4x+7=6x-5

addition property so 4x+12=6x

then subtraction property so 2x=12

divison property x=6

All are equality

ella [17]3 years ago

3 0

Answer:

x = 6

Step-by-step explanation:

4x + 7 = 6x - 5

subtract 7 from both sides (Subtraction Property)

4x = 6x -12

subtract 6x from both sides (Subtraction Property)

-2x = -12

divide by -2 (Division Property)

x = 6

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3 years ago

Amira is painting a rectangular banner 2 1/4 yards wide on a wall in the cafeteria.

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4 years ago

A number is chosen at random from the first 100 positive integers. find the probability that the number is divisible by 5. 1/10

Arlecino [84]

The probability that the number chosen is divisible by 5 is; 1/5.

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2 years ago

Suppose twenty-two communities have an average of = 123.6 reported cases of larceny per year. assume that σ is known to be 36.8

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We are given the following data:

Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22

We are to find the confidence intervals for 90%, 95% and 98% confidence level.

Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.

Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.645* \frac{36.8}{ \sqrt{22}}=110.69$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.645* \frac{36.8}{ \sqrt{22}}=136.51$

Thus the 90% confidence interval will be (110.69, 136.51)

Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.96* \frac{36.8}{ \sqrt{22}}=108.22$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.96* \frac{36.8}{ \sqrt{22}}=138.98$

Thus the 95% confidence interval will be (108.22, 138.98)

Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-2.327* \frac{36.8}{ \sqrt{22}}=105.34$
Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+2.327* \frac{36.8}{ \sqrt{22}}=141.86$

Thus the 98% confidence interval will be (105.34, 141.86)

Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)

As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.

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4 years ago

What is the sum of 9/10+8/100

ArbitrLikvidat [17]

9/10 + 8/100

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Common denominator is 50.

9/10 + 2/25.

5×9/5×10 + 2×2/2×25

45/50 + 4/50

= 49/50

6 0

3 years ago

Read 2 more answers