The width used for the car spaces are taken as a multiples of the width of
the compact car spaces.
Correct response:
- The store owners are incorrect
<h3 /><h3>Methods used to obtain the above response</h3>
Let <em>x</em><em> </em>represent the width of the cars parked compact, and let a·x represent the width of cars parked in full size spaces.
We have;
Initial space occupied = 10·x + 12·(a·x) = x·(10 + 12·a)
New space design = 16·x + 9×(a·x) = x·(16 + 9·a)
When the dimensions of the initial and new arrangement are equal, we have;
10 + 12·a = 16 + 9·a
12·a - 9·a = 16 - 10 = 6
3·a = 6
a = 6 ÷ 3 = 2
a = 2
Whereby the factor <em>a</em> < 2, such that the width of the full size space is less than twice the width of the compact spaces, by testing, we have;
10 + 12·a < 16 + 9·a
Which gives;
x·(10 + 12·a) < x·(16 + 9·a)
Therefore;
The initial total car park space is less than the space required for 16
compact spaces and 9 full size spaces, therefore; the store owners are
incorrect.
Learn more about writing expressions here:
brainly.com/question/551090
The fourth graders raised 175 more than 1,000 for the local animal shelter.
First, let us identify the given.
- Target value to raise= 1,000
- Profit from bake sale= 465
- Profit from t-shirt sales= 710
How much money did the fourth graders raise? The sum of the profit from the bake sale and t-shirt sales determines the total money raised.
Total money= Profit from bake sale + Profit from t-shirt sales
Total money= 465 + 710
Total money= 1175
How much more than 1,000 have the fourth graders raised? This amount is equivalent to the difference between the total money raised and the target amount, 1,000.
Total money-target amount
1175-1000= 175
You can learn more through this link brainly.com/question/24560878.
#SPJ4
Answer:
The range of T is a subspace of W.
Step-by-step explanation:
we have T:V→W
This is a linear transformation from V to W
we are required to prove that the range of T is a subspace of W
0 is a vector in range , u and v are two vectors in range T
T = T(V) = {T(v)║v∈V}
{w∈W≡v∈V such that T(w) = V}
T(0) = T(0ⁿ)
0 is Zero in V
0ⁿ is zero vector in W
T(V) is not an empty subset of W
w₁, w₂ ∈ T(v)
(v₁, v₂ ∈V)
from here we have that
T(v₁) = w₁
T(v₂) = w₂
t(v₁) + t(v₂) = w₁+w₂
v₁,v₂∈V
v₁+v₂∈V
with a scalar ∝
T(∝v) = ∝T(v)
such that
T(∝v) ∈T(v)
so we have that T(v) is a subspace of W. The range of T is a subspace of W.
Answer:
0.3114
Option d is right
Step-by-step explanation:
Let X be the time spent on a treadmill in the health club
Given that research shows that on average, patrons spend an average of 42.5 minutes on the treadmill, with a standard deviation of 5.4 minutes
Also given that X is normal
the probability that randomly selected individual would spent between 30 and 40 minutes on the treadmill.
round off to two decimals tog et
the probability that randomly selected individual would spent between 30 and 40 minutes on the treadmill is 0.31
Hence option d is right
Answer:
m = -6
Step-by-step explanation:
-8m + 4 = 52
-8m = 48
-8m/-8 = 48/-8
m = -6
