All categories

3 years ago

What is 79 1/2% of $7570?

Mathematics

1 answer:

Mariulka [41]3 years ago

5 0

Answer:

$6018.15

Step-by-step explanation:

"is" number always goes on TOP "of" number.

"is" number is 79 1/2%

"of" number is $7570

79 1/2 % / $7570 * 100 = $6018.15 

To find out if the answer is correct :

6018.15 / 7570 * 100 = 79.12%

You might be interested in

Find f^-1(x) and it’s domain.

Mrac [35]

Answer: A

Step-by-step explanation:

Letting $f(y)=x$ ,

$x=\sqrt{y}-5\\\\x+5=\sqrt{y}\\\\y=(x+5)^{2}$

Also, the domain of an inverse is the same as the range of the original function, so the range is $x \geq 0$

7 0

2 years ago

Tyler delivers newspapers.He collected $10 from each customer.He has 5 customers on each street,and delivers to 4 different stre

strojnjashka [21]

Answer:

$200

Step-by-step explanation:

4 different streets with 5 customers on each means 4*5 customers or 20 customers. If he has 20 customers and he gets $10 from each one, thats 20*10. he collected $200

6 0

3 years ago

Read 2 more answers

Graph the system of equations on graph paper.

MrRa [10]

Ths system of equations is

8x+6y=48 ..............>y = 48/6 -8x/6
2x−3y=−6 ..............>y = 6/3 + 2x/3

If we use a graphic tool, we can easily check the solution,
which is (x,y) = (3,4)

The ordered pair lies in Quadrant I

3 0

3 years ago

Read 2 more answers

What simple interest rate is needed for $6,000 to earn $1,000 in 15 months

Mekhanik [1.2K]

let's recall that a year has 12 months, so 15 months is really 15/12 years.

$\bf ~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill&\$1000\\ P=\textit{original amount deposited}\dotfill & \$6000\\ r=rate\to r\%\to \frac{r}{100}\\ t=years\to \frac{15}{12}\dotfill &\frac{5}{4} \end{cases}$

$\bf 1000=(6000)r\left( \cfrac{5}{4} \right)\implies 1000=7500r\implies \cfrac{1000}{7500}=r\implies \cfrac{2}{15}=r \\\\\\ \stackrel{\textit{converting that to a percentage}}{\cfrac{2}{15}\cdot 100 = r\implies \cfrac{40}{3}=r}\implies \stackrel{\%~\hfill }{13.\overline{3}}=r$

4 0

3 years ago

The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78

____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

$m-t_{a/2,n-1}\frac{s}{\sqrt{n} }$ ≤ μ ≤ $m+t_{a/2,n-1}\frac{s}{\sqrt{n} }$

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and $t_{a/2,n-1}$ is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and $t_{0.05,9}$ by 1.8331, we get that the 90% confidence interval for the mean speed is:

$74.2-(1.8331)\frac{5.3083}{\sqrt{10} }$ ≤ μ ≤ $74.2+(1.8331)\frac{5.3083}{\sqrt{10} }$

74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077

71.123 ≤ μ ≤ 77.277

8 0

2 years ago