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andrew11 [14]
3 years ago
8

Kyle invests $11200 in two different accounts. The first account paid 10 %, the second account paid

Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
4 0

Answer:

Kyle invested 5,800 at 10% and 5,400 at 7%.

Step-by-step explanation:

Let x be the amount Kyle invests in the 10% account.

0.1x + 0.07(11200 - x) = 958

0.1x + 784 - 0.07x = 958

0.03x = 174

x = 5800

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The product of two consecutive even positive integers is one hundred twenty. fined the integers
AfilCa [17]
Product is multiplication:

x * (x+2) = 201

x^2 +2x = 120

x^2 + 2x -120 = 0

(x+12) (x-10) = 0

x = -12 and x = 10

-12 x -1 = 12

 the 2 numbers are 10 and 12

3 0
3 years ago
X Y
goldfiish [28.3K]
Range R={-3-11,1,17}
domain D={-9,-3,5}
4 0
3 years ago
please help! find the scale factor and ratio of perimeters for a pair of similar octagons with areas 64ftsq and 81ftsq. i need t
kati45 [8]

Answer:

the scale factor and the ratio of their perimeters is 8/9

Step-by-step explanation:

k^2=area 1/area 2

k^2=64/81

=square root if k^2=square root of 64/81

=<u>k</u><u>=</u><u>8</u><u>/</u><u>9</u>

4 0
2 years ago
Suppose each of the following data sets is a simple random sample from some population. For each dataset, make a normal QQ plot.
adell [148]

Answer:

a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.

c) This last case the histogram is not symmetrical and the data seems to be skewed.

Step-by-step explanation:

For this case we have the following data:

(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

We can use the following R code to get the histogram

> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

> hist(x1,main="Histogram a)")

The result is on the first figure attached.

For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> hist(x2,main="Histogram b)")

The result is on the first figure attached.

For this case the data is skewed to the left and we can't assume that we have the normality assumption.

(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> hist(x3,main="Histogram c)")

The result is on the first figure attached.

This last case the histogram is not symmetrical and the data seems to be skewed.

7 0
3 years ago
What percent of 78 is 16?
Phantasy [73]

Answer:

78/16=

39/8=

19.5/4= 4.80805

7 0
3 years ago
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