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HACTEHA [7]
3 years ago
12

1. Find the distance and midpoint of XY when X(-7, 10) and Y(3, 4).

Mathematics
2 answers:
nordsb [41]3 years ago
6 0
I don’t know the answer
lutik1710 [3]3 years ago
5 0
Distance:

d = √(X2 - X1)² + (Y2 - Y1)²
d = √(3 -(-7))²+(4-10)²
d = √(3+7)²+(4-10)²
d = √(10)²+(-6)²
d = √100+36
d = √136
d ≈ 11.66
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Paul says that (3.14 X 10^5) + (2.53 X 10^4) = 5.67 X 10^5. Is Paul correct? Explain.
Marizza181 [45]
He is false the equation should be as follows

3 0
3 years ago
Christine,milan, and Scott sent a total of 145 messages during the weekend . Christine sent 7 more messages than Scott . Milan s
AURORKA [14]

Answer:

number of message sent by Scott is 23

number of message sent by Christine is   30

number of message sent by Milan is 92

Step-by-step explanation:

Total message sent by Christine,Milan, and Scott : 145

Let number of message sent by Scott be x .

Given that Christine sent 7 more messages than Scott

number of message sent by Christine is   x+7

It is also given that  Milan sent 4 times as many messages as Scott

number of message sent by Milan is   4*x = 4x.

Thus, sum of message sent by Christine,milan, and Scott  in terms of x

= x + x + 7 + 4x = 6x + 7

we already know that Total message sent by Christine,milan, and Scott is 145

thus,  6x + 7 should be equal to 145

6x + 7 = 145

=>6x = 145 - 7 = 138

=> x = 138/6 = 23

Thus,  

number of message sent by Scott is x = 23

number of message sent by Christine is   x+7 = 23+7 = 30

number of message sent by Milan is   4x = 4*23 = 92

3 0
4 years ago
PLEASE ANSWER FAST!! will mark brainiest is answered correctly.
Olenka [21]
Number 1 is 4 4/5#2 is-3 #3 is > anf # 4 is 13 units
5 0
4 years ago
Three different methods for assembling a product were proposed by an industrial engineer. To investigate the number of units ass
In-s [12.5K]

Answer:

is an attachment

test statistic = 9.87

p value is less than 0.01

Not all means of the three assembly methods are equal

Step-by-step explanation:

the anova table is an attachment

total number of methods = k = 3

number of observations n = 30

for treatment, df = 3-1 = 2

for error, df = 30 -3 (n-k) = 27

sst = 10800, sstr = 4560, sse = 10800-4560 = 6240

we find the mean of squares for error

sse/df of error = 6240/27 = 231.11

mean of squares for treatment = 4560/2 = 2280

test stat

F = 2280/231.11

= 9.8654

using f distribution table;

alpha = 0.05

df = 2 , 27

we get 3.35

h0; no difference in mean

h1; there is difference

9.87 > 3.35 so we reject H0, there is difference in means. all are not equal.

pvalue calculation

using excel,

FDIST(9.8654, 2, 27) = 0.0006078

p value is less than 0.01

we conclude that a. Not all means of the three assembly methods are equal

3 0
3 years ago
Solve the proportion
weeeeeb [17]

Answer:

t = 3/16

Step-by-step explanation:

Solve the proportion fraction numerator 15 t over denominator 5 end fraction equals fraction numerator 2 t plus 3 over denominator 6 end fraction

From the above question, we are to solve for t

The above expression is written Mathematically as:

15t/5 = 2t + 3/6

We cross Multiply

15t(6) = 5(2t + 3)

90t = 10t + 15

Collect like terms

90t - 10t = 15

80t = 15

t = 15/80

t = 3/16

7 0
3 years ago
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