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marusya05 [52]
1 year ago
7

you are planning to purchase displays for your store which has 2400 sq ft of selling space. Each display has a footprint of 10 f

t x 4 ft. You want to leave 60% open space in your store for walking around. How many displays can you fit?
Mathematics
1 answer:
Fynjy0 [20]1 year ago
5 0

60% of 2,400 is 1,440. So in all actuallity, you only have 960 square feet to fill with displays. 10*4= 40ft^2. 960/40=24.

You can fit 24 displays in your store while having 60% walking space.

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Rhianna is buying a car for $14,390. She has a $1000 trade-in allowance and will make a $1500 down payment. She will finance the
antoniya [11.8K]

Answer:

(a)

Original price for the Car Rhianna buy = $14,390.

Trade in allowance she has =$1000

So, the reduced price= $14,390-$1,000=$13,390.

Now, she make a $1500 down payment initially, so the remaining amount she pay using Auto-Loan is,    $13,390-$1500=$11,890

Therefore, $11,890 money will she borrow in an auto loan.

(b)

To find the monthly(M) auto payment be:

Since the monthly auto payment in 4 year with 2.6% APR(Annual percentage rate)

Now, Monthly Auto payment(M) be :

P=$11,890, Monthly Car payment per $1000 borrowed at 2.6% for 48 months = $21.958

∴ M=$11,890\cdot \frac{21.958}{1000}

On solving we get,  M=$261.082

Therefore, $261.082 will be her monthly auto payment.

(c)

Total amount of interest(A) she will pay ,

A=total payment in 4 years-total loan amount=(\$261.082\cdot 48)-11,890

=$12,531.936-$11,890=$641.936

So, the total amount of interest she will pay is, $641.936

(d)

Rhianna total payment for the car = total payment in 4 years+Down payment=(261.082\cdot 48)+$1500=$12531.936+$1500=$14,031.936

(e)

∵ Rhianna is 22 years female :

From the given Data in the picture:

Rating factor= 1.2  

Liability insurance 50/100/25 , Bodily injury 50/100

Premium= $310 and Property damage 25 : Premium $175  

Collision insurance=$500 deductible:  Premium $148  

Comprehensive insurance= $500 deductible : Premium $85  

Her total annual premium= 1.2\cdot (\$310+\$175+\$148+\$85) = \$861.60

8 0
1 year ago
If 25 dimes were moved from Box A to Box B, there would be an equal amount of dimes in both boxes. If 100 dimes were moved from
nexus9112 [7]
Ok.look.the question is like that
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5 0
2 years ago
If f(x)=x2+10sinx, show that there is a number c such that f(c)=1000.
gayaneshka [121]
F(x) is continuous for all x.

Pick a point and show that f(x) is either negative or positive. Pick another point and show that f(x) is negative, if positive, or positive, if negative.

At x = 30, f(30) - 1000 = 900 + 10sin(30) - 1000 ≤ 0
Now, show at another point f(x) - 1000 is positive, and hence, there would be root between 30 and such point.

Let's pick 40.
At x = 40, f(40) - 1000 = 1600 + 10sin(40) - 1000 ≥ 0

Since f(x) - 1000 is continuous, there lies a root between 30 and 40, and hence, 30 ≤ c ≤ 40
7 0
2 years ago
In isosceles △ABC (AC = BC) with base angle 30° CD is a median. How long is the leg of △ABC, if sum of the perimeters of △ACD an
SIZIF [17.4K]

Note necessary facts about isosceles triangle ABC:

  • The median CD drawn to the base AB is also an altitude to tha base in isosceles triangle (CD⊥AB). This gives you that triangles ACD and BCD are congruent right triangles with hypotenuses AC and BC, respectively.
  • The legs AB and BC of isosceles triangle ABC are congruent, AC=BC.
  • Angles at the base AB are congruent, m∠A=m∠B=30°.

1. Consider right triangle ACD. The adjacent angle to the leg AD is 30°, so the hypotenuse AC is twice the opposite leg CD to the angle A.

AC=2CD.

2. Consider right triangle BCD. The adjacent angle to the leg BD is 30°, so the hypotenuse BC is twice the opposite leg CD to the angle B.

BC=2CD.

3. Find the perimeters of triangles ACD, BCD and ABC:

P_{ACD}=AC+CD+AD=2CD+CD+AD=3CD+AD;

P_{BCD}=BC+CD+BD=2CD+CD+AD=3CD+AD;

P_{ABC}=AC+BC+AB=2CD+2CD+AD+BD=4CD+2AD.

4.  If sum of the perimeters of △ACD and △BCD is 20 cm more than the perimeter of △ABC, then

P_{ACD}+P_{BCD}=P_{ABC}+20,\\ \\3CD+AD+3CD+AD=4CD+2AD+20,\\ \\6CD+2AD=4CD+2AD+20,\\ \\2CD=20.

5. Since AC=BC=2CD, then the legs AC and BC of isosceles triangles have length 20 cm.

Answer: 20 cm.

8 0
2 years ago
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earnstyle [38]
It is 1.5 times because $63 divided by $42 is $1.5.
I hope this is what you meant. 
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2 years ago
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