Answer:
i think it's <C and <R since they're 90 degree angles, though they are small
To write this as a percent, just move the decimal two places to the right. So instead of .098, this becomes 9.8%. hope this helps!
Answer:
Hope it helps u
Step-by-step explanation:
As we know that ,
Mean = sum of the terms/ numbers of terms
But here grouped data is given so , we use the formula
Mean=∑[f. m]/ ∑f
where f is frequency and m is mid point of each height ,
Now first we have to find the mid point of each interval, where
midpoint of each interval = (lower boundary + upper boundary)/2
m1=(150+154)/2 = 152
m2=(155+159)/2= 157,now found other by same formula, for each interval
m3= 162
m4= 167
m5=172 Now we find the midpoint of each interval ,so now
∑[f. m]=f1*m1+f2*m2+f3*m3+f4*m4+f5*m5
now putting the values of each frequency for given interval and midpoint of each interval we will get,
∑[f. m]=456+942+1296+167*x+344 = 167*x+3038
Now find,
∑f=f1+f2+f3+f4+f5
∑f=19+x
Now we have,
∑[f. m]=167*x+3038
∑f=19+x
also given mean height=161.6 cm
putt these values in above equation we get,
161.6=
now solve this ,
161.6(19+x)=167*x+3038
3070.4+161.6*x=167*x+3038
3070.4-3038=167*x-161.6*x
32.4=5.4*x
x=32.4/5.4
<h2>
x=6 Ans........</h2>
In the given question, there are two vital information's give. Firstly 8 ears of corn and 1 cantaloupe cost $2.37. On the other hand 6 ears of corn and 3 cantaloupes cost $3.51.
Let us assume the cost of 1 corn = X
Let us assume the cost of 1 cantaloupe = Y
Then
8X + Y = 2.37
Y = 2.37 - 8X
And
6X + 3Y = 3.51
Now we replace the value of Y we found from the first equation
Then
6X + 3(2.37 - 8X) = 3.51
6X - 24X + 7.11 = 3.51
-18X = 3.51 - 7.11
-18X = -3.6
X = 3.6/18
= 0.20
So $0.20 is the value of 1 corn. Then the value of a cantaloupe
Y = 2.37 - 8X
= 2.37 - (8 * 0.20)
= 2.37 - 1.6
= 0.77
So cost of a cantaloupe is $0.77
Answer:
5130
Step-by-step explanation:
You just multiply 513 by 10 to get 5130.
You can check this answer by just seeing if it fits the requirements, which it does. Making it the correct answer.