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netineya [11]
3 years ago
11

Using only nickels, dimes and quarters, the minimum number of coins needed to cover every possible choice of 5¢, 10¢, 15€, --- t

o $1.35 is
Mathematics
1 answer:
creativ13 [48]3 years ago
6 0

Answer:

6 coins 5 quarters and one nickel.

Step-by-step explanation:

5 quarters = 1.25 + 1 nickel = .10  

(1.25 + .10 = 1.35)  hope it helps.

                                                         :)

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If the sales tax rate is 7.25% in California and 20% gratuity is automatically added to your restaurant check. How much would yo
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The tax added to $220 is solved by 220 X .0725 = 15.95. Add 15.95 to $220 and get $235.95. The tip (assuming it is 20% after tax ) is $235.95 X .20= $47.19
Add the $47.19 to $235.95= $283.14
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BRAINLIEST FOR QUICKEST<br> Please help with this question!!
GarryVolchara [31]

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A and B and C and D

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3 years ago
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How to do this and answer this
nordsb [41]

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

                  Hello!

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

❖ 8² - (4+2)² × 2 ÷ 4 = 46

Follow order of operations:

Parenthesis

Exponents

Multiplication

Division

Addition

Subtraction

(Note: Multiplication and division can be done in either order as well as addition and subtraction.)

First, do parenthesis and exponents:

(4 + 2)² = 36   8² = 64

Now do multiplication and division:

36 x 2 ÷ 4 = 18

Now do subtraction:

64 - 18 = 46

~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

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7 0
3 years ago
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Each day a commuter takes a bus to work, the transportation system has a phone app that tells her what time the bus will arrive.
Paraphin [41]

Answer:

Step-by-step explanation:

Hello!

The commuter is interested in testing if the arrival time showed in the phone app is the same, or similar to the arrival time in real life.

For this, she piked 24 random times for 6 weeks and measured the difference between the actual arrival time and the app estimated time.

The established variable has a normal distribution with a standard deviation of σ= 2 min.

From the taken sample an average time difference of X[bar]= 0.77 was obtained.

If the app is correct, the true mean should be around cero, symbolically: μ=0

a. The hypotheses are:

H₀:μ=0

H₁:μ≠0

b. This test is a one-sample test for the population mean. To be able to do it you need the study variable to be at least normal. It is informed in the test that the population is normal, so the variable "difference between actual arrival time and estimated arrival time" has a normal distribution and the population variance is known, so you can conduct the test using the standard normal distribution.

c.

Z_{H_0}= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } }

Z_{H_0}= \frac{0.77-0}{\frac{2}{\sqrt{24} } }= 1.89

d. This hypothesis test is two-tailed and so is the p-value.

p-value: P(Z≤-1.89)+P(Z≥1.89)= P(Z≤-1.89)+(1 - P(Z≤1.89))= 0.029 + (1 - 0.971)= 0.058

e. 90% CI

Z_{1-\alpha /2}= Z_{0.95}= 1.645

X[bar] ± Z_{1-\alpha /2}* (\frac{Sigma}{\sqrt{n} } )

0.77 ± 1.645 * (\frac{2}{\sqrt{24} } )

[0.098;1.442]

I hope this helps!

4 0
3 years ago
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