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FromTheMoon [43]
3 years ago
13

Compute the total cost per year of the following pair of expenses. Then complete the​ sentence: On an annual​ basis, the first s

et of expenses is​ _______% of the second set of expenses.
Vern buys seven lottery tickets each week at a cost of ​$3 each and spends ​$900 per year on his textbooks.
Mathematics
1 answer:
Mandarinka [93]3 years ago
6 0

Answer:

So 912$ is 58% of 1584 $

Step-by-step explanation:

You might be interested in
Hlp me with this math problem
zaharov [31]

Answer:

0.5l+0.2s

Step-by-step explanation:

0.10x5<em>l</em> equals 0.5<em>l</em>

0.10x2<em>s</em> equals 0.2<em>s</em>

7 0
3 years ago
F(x) = -x + 4 and g(x) = 6x + 3, find (f(x) - g(x))<br><br>A. -5x-1<br>B. 5x +1<br>C. -7x +1<br>​
Pepsi [2]

Answer:

C

Step-by-step explanation:

f(x) - g(x)

= - x + 4 - (6x + 3)

= - x + 4 - 6x - 3 ← collect like terms

= - 7x + 1 → C

8 0
3 years ago
Select all ratios equivalent to 3:5.<br> 13:30<br> 4:15<br> 6:10
uysha [10]

Answer:

6:10

Step-by-step explanation:

3:5 is equivalent to 6:10 because you multiply both 3 and 5 by 2 and you get 6:10. every thing else is wrong

6 0
2 years ago
Read 2 more answers
Use the following figure to find the value of x <br>(plz help)​
masha68 [24]
X-6=1/3x( solve equation)
x-1/3x=6
0.7x=6
x=6/0.7
x=8.5 ( i guess its like this)
7 0
3 years ago
If the distribution is really (5.43,0.54)
defon

Answer:

0.7486 = 74.86% observations would be less than 5.79

Step-by-step explanation:

I suppose there was a small typing mistake, so i am going to use the distribution as N (5.43,0.54)

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

The general format of the normal distribution is:

N(mean, standard deviation)

Which means that:

\mu = 5.43, \sigma = 0.54

What proportion of observations would be less than 5.79?

This is the pvalue of Z when X = 5.79. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{5.79 - 5.43}{0.54}

Z = 0.67

Z = 0.67 has a pvalue of 0.7486

0.7486 = 74.86% observations would be less than 5.79

7 0
3 years ago
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