Use the distributive property to write an expression equivalent to 5×(3+4)

Determine the slope of the line passing through the given points (1,2) and (2,-1)

alex41 [277]

I hope this helps you

slope=y2-y1/x2-x1

slope=2-(-1)/1-2

slope=3/-1

slope= -3

3 0

3 years ago

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Theresa and Mark are playing cards. Each card is -50 points.

Amiraneli [1.4K]

*Since each card is worth -50 points, you need to remember that a positive times a negative integer gives you a negative integer.*

Part A) Theresa has 5 cards so: 5 x -50 = -250

Theresa has -250 points.

Part B) Mark has 8 cards so: 8 x -50 = -400

Mark has -400 points.

Part C) Theresa has more points because -250 is closer to 0 on a number line.

4 0

3 years ago

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Solve the following dividend problem.

brilliants [131]

A.

To get dividend per share, we need to divide total dividend by total numbers of shares outstanding.

Dividend per share = $\frac{40,000}{50,000}=0.8$ dollars.

B.

Since per share you get $0.8, for 100 shares you will get $0.8*100=80$ dollars.

Also, that is $\frac{80}{40,000}*100=0.2$ percent of the total dividends.

ANSWER:

A. $0.8 per share

B. $80 (and that is 0.2% of total dividend)

7 0

3 years ago

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A box of 6 glue sticks costs $8.95. How much will it cost to buy 5 glue sticks?

ycow [4]

It will cost you $7.45

4 0

2 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

2 years ago

Use the distributive property to write an expression equivalent to 5×(3+4)​

Use the distributive property to write an expression equivalent to 5×(3+4)