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2 years ago

Ok I’m sorry but this is the last question. I need help please

Mathematics

1 answer:

Lynna [10]2 years ago

7 0

Answer:

19%

Step-by-step explanation:

The first discount is the 10% off.

An item costs 100% of its original price. By discounting 10%, the discounted price is now only 90% of the original price.

The price is now 90% of the original price. This is before the coupon is used.

Now we use the coupon. The coupon is 10% off of the discounted price.

That means now you will pay 90% of the discounted price.

The discounted price is 90% of the original price, so after the 10% coupon, you pay 90% of 90% of the original price.

90% * 90% = 0.90 * 0.90 = 0.81 = 81%

After the discount and the coupon, you pay 81% of the original price.

The price before the discount and the coupon was 100% of the original price.

Now the price is 81% of the original price.

100% - 81% = 19%

The final price has a total percent of discount of 19%.

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Which equation represents f(x)?

Marta_Voda [28]

Answer:

=which*equation*represents*f((x))

the*graph*of*the*cube*root*parent*function*y=v*is

=translated*to*form*f((x))shown*on*the*graph*0.

f((x))=+6+1

f((x))=x-6+1

f((x))=x+6-1

f((x))=3x-6-1

=10

e^(x)=2

==========

x = 0.693115234375

7 0

2 years ago

Don't stick with your computer, do smth that make you happy IRL<br> btw, don't report this please

Mrac [35]

What makes me happy irl is nothing really, I only exist to make others happy

5 0

2 years ago

Read 2 more answers

(1) a) 5270 cl= ? dal

Ivanshal [37]

Answer:

Check below

Step-by-step explanation:

1) These metric volume units can be easily converted by dividing or multiplying by 10 and its multiple. Like this: each step up on the ladder multiply by 10. Each step down divide by 10

$a)5270 \:cl=527\:dl\\b)kl=5.3 \times 10=530 hl\\c)58000ml=5.8 dal =(\frac{58000}{10000})\\d)0.39 dal=3.9 l\\e)7.8 dal=780 dl\\f)57.2 ml=5.72 cl$ .

2) When it comes to area, the "ladder scheme" remains valid but now we'll multiply or divide by

$10^3$

Bear in mind these useful relations:

$1litre =\frac{1}{10^3}m^{3}$

$1 cm^{3} =1 ml$

$a) 3.7 l=3700 cm^{3}\\b)29.6 dm^{3}\\29.6 dm^{3} \rightarrow 29.6\times10^{3} cm^3=29.6\times10^{3} ml = 29.6\times10^{2} cl \:or\:2960 cl\\c)8.1l=8100 cm^{3}\\d)5l=\frac{5}{10^{-3}} m^{3}\\e)5.2 cl=52ml=52 cm^{3}=52\times10^{3}=52000 \:mm^{3}\\f)375 ml=375 cm^{3}\\g) 1300 cm^{3}=1300 ml=13 dl\\h) 8.6 ml= 8.6 cm^{3}\times\: 10^{3}=8600 mm^{3}\\i) 51.8\times 10^{11} l \rightarrow 51.8\times 10^{-11}km^{3}$

6 0

2 years ago

Among freshman at a certain university, scores on the Math SAT followed the normal curve, with an average of 550 and an SD of 10

lina2011 [118]

Answer:

a) 6.68th percentile

b) 617.5 points

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 550, \sigma = 100$

a) A student who scored 400 on the Math SAT was at the ______ th percentile of the score distribution.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{400 - 550}{100}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

So this student is in the 6.68th percentile.

b) To be at the 75th percentile of the distribution, a student needed a score of about ______ points on the Math SAT.

He needs a score of X when Z has a pvalue of 0.75. So X when Z = 0.675.

$Z = \frac{X - \mu}{\sigma}$

$0.675 = \frac{X - 550}{100}$

$X - 550 = 0.675*100$

$X = 617.5$

6 0

2 years ago

As his empire spread throughout the ancient world, Alexander founded roughly how many cities?

Studentka2010 [4]

About 70 Cities in Total.

For his more noteworthy grandness, Alexander established exactly 70 urban areas in the grounds he vanquished and requested them named after himself. Most celebrated, obviously, is Alexandria in Egypt. In India, when his dearest horse kicked the bucket, he requested a city to be constructed named Bucephala.

6 0

3 years ago