Carrots: 50/100
Broccoli: 25/100
Peas: 10/100
Celery: 15/100
Z score ( X = 1890 ): 1.31
Z score ( X = 1230 ): -0.76
Z score ( X = 2220 ): 2.34 (This value of Z is unusual )
Z score ( X = 1360 ): -0.35
<u>Step-by-step explanation:</u>
Here we have , A standardized exam's scores are normally distributed. In a recent year, the mean test score was 1473 and the standard deviation was 318 . The test scores of four students selected at random are 1890 , 1230 , 2220 , and 1360 . We need to find Find the z-scores that correspond to each value and determine whether any of the values are unusual. Let's find out:
We know that Z score is given by : ( data - Mean ) / ( standard deviation )
Z score ( X = 1890 ):
⇒
Z score ( X = 1230 ):
⇒
Z score ( X = 2220 ):
⇒
This value of Z is unusual as Value lies as : .
Z score ( X = 1360 ):
⇒
Answer:
4a + 3b
Step-by-step explanation:
Mark brainliest
Answer:
4%
Step-by-step explanation:
New value:
$
71.50
New value: $71.50
Old value:
71.5
−
2.75
=
$
68.75
Old value: 71.5−2.75=$68.75
68.75
(
1
+
r
)
=
71.5
68.75(1+r)=71.5
68.75
(
1
+
r
)
68.75
=
71.5
68.75
68.75
68.75(1+r)
=
68.75
71.5
1
+
r
=
1.04
1+r=1.04
r
=
0.04
r=0.04
Subtract 1
The stock gained 4
%
The stock gained 4%
Multiply by 100
Answer:
a)12 b)12 c)36
Step-by-step explanation:
a)find least common multiple. the number of marbles is that number * n (where n is an integer)
b) the least common multiple. it is 4*3=...
c)12*3 =?
then just divide by 2,3,4 to get answer