All categories

3 years ago

A jar contains 11 pennies and 8 dimes. What is the ratio of dimes to pennies?

Mathematics

2 answers:

Galina-37 [17]3 years ago

6 0

I hope this helps you

dimes = 8

pennies = 11

8:11

Trava [24]3 years ago

3 0

Dimes to pennies = 8 to 11

You might be interested in

60 out of 300 CDs are classical music CDs. What percent of the CDs are classical music CDs?

AnnyKZ [126]

300 divided by 60 is 5 so that mean 60 goes into 300 5 times so each time is 20 %

7 0

3 years ago

Read 2 more answers

The Candela brothers own two pizza restaurants, one on Park Street and one on Bridge Road.

koban [17]

The mean, median and mode are measures of central tendency, that is they tend to indicate the location middle of the data

Required values;

(a) The performance for the week for Park Street

Revenue is Q₂ < $7,500 < Q₃

The sales for the week is better than 72.91% of all sales

The performance for the week for Bridge Road

Revenue; Q₂ < $7,100 < Q₃

The sale for the week is better than 59.87% of all sales

(b) The mean is $3611

The median is $3,600

The standard deviation is $3250

The Interquartile range is $6075

Reason:

The table of values that maybe used to find a solution to the question is given as follows;

$\begin{array}{|l|l|l|}\mathbf{Variable} &\mathbf{Park}&\mathbf{Bridge}\\N&36&40\\Mean&6611&5989\\SE \ Mean&597&299\\StDev&3580&1794\\Minimum&800&1800\\Q_1&3600&5225\\Median&6600&6000\\Q_3&9675&7625\\Maximum&14100&8600\end{array}\right]$

(a) Park Street revenue = $7,500

Bridge Road's revenue = $7,100

The two stores sold close to but below the 75th percentile

Bridge Road revenue;

The z-score is given as follows;

$Z = \dfrac{x - \mu }{\sigma }$

$Z = \dfrac{7100 - 5,989 }{1794 } \approx 0.6193$

From the Z-Table, we have;

The percentile= 0.7291

Therefore, the sale for the week for Park Street is better than 72.91% of all the sales

Park Street revenue;

The z-score is given as follows;

$Z = \dfrac{7500 - 6611}{3580} \approx 0.25$

From the Z-Table, we have;

The percentile = 0.5987

Therefore, the sale for the week is better than 59.87 % of all the sales

(b) Given that the operating cost is $3,000, frim which we have;

The subtracted value is subtracted from the mean and median to find the new value

Profit = The revenue - Cost

New mean = 6611 - 3000 = 3611

The new mean = $3,611

The new median = 6600 - 3000 = 3600

The new median = $3,600

The standard deviation and the interquartile range remain the same, therefore, we have;

The standard deviation = $3,580

The interquartile range = 9675 - 3600 = 6075

The interquartile range = 6075

Learn more here:

brainly.com/question/21133077

brainly.com/question/23305909

5 0

1 year ago

What is the equivalent factored form of 12x^4-42x^3-90x^2

Tanzania [10]

Factor the following:
12 x^4 - 42 x^3 - 90 x^2

Factor 6 x^2 out of 12 x^4 - 42 x^3 - 90 x^2:
6 x^2 (2 x^2 - 7 x - 15)

Factor the quadratic 2 x^2 - 7 x - 15. The coefficient of x^2 is 2 and the constant term is -15. The product of 2 and -15 is -30. The factors of -30 which sum to -7 are 3 and -10. So 2 x^2 - 7 x - 15 = 2 x^2 - 10 x + 3 x - 15 = x (2 x + 3) - 5 (2 x + 3):
6 x^2 x (2 x + 3) - 5 (2 x + 3)

Factor 2 x + 3 from x (2 x + 3) - 5 (2 x + 3):
Answer: 6 x^2 (2 x + 3) (x - 5)

4 0

3 years ago

Read 2 more answers

Find the mean, variance &a standard deviation of the binomial distribution with the given values of n and p.

MrMuchimi

A random variable following a binomial distribution over $n$ trials with success probability $p$ has PMF

$f_X(x)=\dbinom nxp^x(1-p)^{n-x}$

Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.

$\displaystyle\sum_xf_X(x)=\sum_{x=0}^n\binom nxp^x(1-p)^{n-x}=1$

The mean is given by the expected value of the distribution,

$\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^nx\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^n\frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\frac{(n-1)!}{x!((n-1)-x)!}p^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\binom{n-1}xp^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^{n-1}\binom{n-1}xp^x(1-p)^{(n-1)-x}$

The remaining sum has a summand which is the PMF of yet another binomial distribution with $n-1$ trials and the same success probability, so the sum is 1 and you're left with

$\mathbb E(x)=np=126\times0.27=34.02$

You can similarly derive the variance by computing $\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$ , but I'll leave that as an exercise for you. You would find that $\mathbb V(X)=np(1-p)$ , so the variance here would be

$\mathbb V(X)=125\times0.27\times0.73=24.8346$

The standard deviation is just the square root of the variance, which is

$\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834$

7 0

3 years ago

HELP THIS is due in 30 mins Renaldo described the translation of the graph of f(x) = x2 related to g(x) = (x + 2)2 - 6 as 2 unit

stiv31 [10]

Answer:

See below

Step-by-step explanation:

Recall that we need to use the equation y=a(x-h)^2+k. This means that h=-2 instead of 2 otherwise it would've been (x-2)^2-6. So Renaldo made the mistake of identifying h as 2 instead of -2. The second mistake Renaldo made is that since k=-6, there should be a vertical shift of 6 units down not 2.

5 0

3 years ago