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Dafna1 [17]
3 years ago
6

PLZ HELP NOW What is the greatest common factor of 10 and 2?

Mathematics
2 answers:
Digiron [165]3 years ago
7 0

Answer:

2

Step-by-step explanation:

lesya [120]3 years ago
5 0

Answer:

The asnwer is 2

uh your welcome

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4567 divided by 13 please <br> help
katrin2010 [14]
351.3 you can use calculators you know lol
6 0
3 years ago
Please help me! Suppose Sal's total profit on lunch specials for the next month is $1,593. The profit amounts are the same: $2 f
blsea [12.9K]

Answer:

Step-by-step explanation:

This is an incomplete problem. Other data were not given.

Given:

Profit of every sandwich = $2

Profit of every wrap = $3

x = sandwich

y = wrap

Last month:  2x + 3y = 1,470

Next month: 2x + 3y = 1,593

Based on the given equation:

Both still have the same profit. $2 for sandwiches and $3 for wraps.

The only reason why there is a difference in the total amount is the change in the number of sandwich or wrap sold in a given month.

Since, next month's total sale is higher than last month's total sale, it is safe to assume that the sale of sandwich or wrap is higher than last month's sale.

4 0
3 years ago
Write the equation of the line given the slope 2 and one point on the line (2, -4)
ELEN [110]
Point Slope Form:
y - ( - 4) = 2(x - 2)
Slope-intercept Form:
y = 2x
6 0
4 years ago
46. Melanie wants to construct the perpendicular bisector of
vekshin1

Answer:

C

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
The accompanying data represent the miles per gallon of a random sample of cars with a​ three-cylinder, 1.0 liter engine.a. Comp
Kobotan [32]

Answer:

1). Z score = 1.464733

Mean = 38.87917

Standard deviation = 3.609405

2). Quartiles:

Q1 = 37.025

Q2 = 38.450

Q3 = 40.800

3). IQR = 3.775

4).

CI (lower fence) = 37.4351

CI (upper fence) =  40.32323

5). There is outlier in the data set. Please see attached box plot for evidence.

Step-by-step explanation:

1). By Z score, we mean:

Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}},

where:

x-bar ==> mean(g) = 38.87917

\mu = 37.80

standard deviation = 3.609405

sample size (n) = 24.

2). By quantile, we mean:

Q = L + (i*(n/4) - Cf)*c

Where L is the lower class limit of the quartile class

Cf is the cumulative frequency before the quartile class

c  is the class size.

3) . IQR = Q3 - Q1

4). CI = \mu \pm *Z_{\alpha/2} \frac{\sigma}{\sqrt{n}},

Where Z_{\alpha/2}  ==> 1.96

In order to replicate and obtain the result, please use the R code below:

g = c(31.5, 36.0, 37.8, 38.5, 40.1, 42.2,34.2, 36.2, 38.1, 38.7, 40.6, 42.5,34.7, 37.3, 38.2, 39.5,  

41.4, 43.4,35.6, 37.6, 38.4, 39.6, 41.7, 49.3)

boxplot(g)

Z = (mean(g) - 37.8)/(sd(g)/sqrt(length(g)))

mean(g)

quantile(g)

IQR(g)

CIl = mean(g) - 1.96*(sd(g)/sqrt(length(g)))

CIU = mean(g) + 1.96*(sd(g)/sqrt(length(g)))

4 0
3 years ago
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