Pls Answer the Question in the Photo

I gave my brother 1/5 of my candy, 1/4 to my friend and had 11 pieces left over. How many pieces did I give my brother?

AnnyKZ [126]

x = total number of candies

$\bf \stackrel{\textit{my brother}}{\cfrac{1}{5}x}+\stackrel{\textit{my friend}}{\cfrac{1}{4}x}+11~~=~~\stackrel{\textit{total}}{x} \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{20}}{20\left( \cfrac{x}{5}+\cfrac{x}{4}+11 \right)=20(x)}\implies 4x+5x+220=20x\implies 9x+220=20x \\\\\\ 220=11x\implies \cfrac{220}{11}=x\implies 20=x~\hspace{10em}\stackrel{\textit{to my brother}}{\cfrac{20}{5}\implies 4}$

4 0

3 years ago

You have 6 donuts and you want to give 2/3 of them to a friend and keep 1/3 for yourself. How many donuts would your friend get?

Furkat [3]

4 donuts because you can change the denominator to 6 and get 4/6 and thats how you get 4

3 0

3 years ago

Read 2 more answers

Can someone explain to how change a percent to a fraction is really confused

stealth61 [152]

For example,
45/100 In decimal would be 0.45
45/1000 in decimal would be 0.045.
I didn't really explain, but I gave an example. Hope this helped you?

6 0

3 years ago

Read 2 more answers

A professor of women's studies is interested in determining if stress affects the menstrual cycle. Ten women are randomly sample

BabaBlast [244]

Answer:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

$t=\frac{(20.4 -27.6)-(0)}{\sqrt{\frac{2.074^2}{5}}+\frac{2.702^2}{5}}=-4.73$

$df=5+5-2=8$

$t_{crit}=\pm 2.306$

Reject H0: stress affects the menstrual cycle

Step-by-step explanation:

The statistic is given by the following formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{s^2_2}{n_2}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 = 0$

Alternative hypothesis: $\mu_1 -\mu_2 \neq 0$

Our notation on this case :

$n_1 =5$ represent the sample size for group 1

$n_2 =2$ represent the sample size for group 2

Tha data given is:

Group 1 (High stress) : 20,23,18,19,22

Group 2 (Relatively stress free): 26,31,25,26,30

We can calculate the sample mean and the sample deviation with the following formulas:

$\bar X =\frac{\sum_{i=1}^n x_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}$

$\bar X_1 =20.4$ represent the sample mean for the group 1

$\bar X_2 =27.6$ represent the sample mean for the group 2

$s_1=2.074$ represent the sample standard deviation for group 1

$s_2=2.702$ represent the sample standard deviation for group 2

And now we can calculate the statistic:

$t=\frac{(20.4 -27.6)-(0)}{\sqrt{\frac{2.074^2}{5}}+\frac{2.702^2}{5}}=-4.73$

Now we can calculate the degrees of freedom given by:

$df=5+5-2=8$

Now we can calculate the critical value since the confidence is 95% the value for the significance would be $\alpha=1-0.95=0.05$ and the value for $\alpha/2 =0.025$ and if we find a critical value on th t distribution with 8 degrees of freedom that accumulates 0.025 of the area on each tail we got $t_{crit}=\pm 2.306$

Since our calculated value is lower than the critical value, we have enough evidence to reject the null hypothesis. And makes sense say that the difference between the two means are different at 5% of significance.

Reject H0: stress affects the menstrual cycle

5 0

3 years ago

An entrepreneur faces many bureaucratic and legal hurdles when starting a new business. The World Bank collects information abou

slamgirl [31]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the data :

Rearranged data:

5, 5, 5, 5, 6, 7, 7, 7, 8, 12, 12, 13, 13, 15, 18, 18, 27, 28, 36, 48, 52, 60, 66, 94

Start time for Suriname = 694

Mean = ΣX / n

Mean = (567 + 694) / (24 + 1) = 1261 / 25 = 50.44

B.) median start Time for initial 24 values

0.5(n+1)th term

0.5(25) = 12.5th term

(13 + 13) / 2 = 13

C.) preferred measure of center for the distribution will be the median as it shapes well in the middle of the distribution. The mean is overestimated

d) Compute the quartiles (Q1, Q2, Q3), and find the IQR for these data. Are there outliers in the time to start a business data set? If so, identify and name any outliers.

Using calculator :

Lower quartile Q1 --> 7

Median Q2 --> 13

Upper quartile Q3 --> 42

IQR = (Q3 - Q1) = 42 - 7 = 35

OUTLIER:

Lower bound : Q1 - 1.5(IQR) ; 7 - 1.5(35) = - 45.5

Upper bound : Q3 + 1.5(IQR) = 42 + 1.5(35) = 94.5

Outlier : values below - 45.5 and above 94.5

Hence, Surimanes start time is the only Outlier.

e)

Standard deviation for 24 countries :

Standard deviation = sqrt[Σ(X - mean)^2 / (N - 1)]

Usibg calculator :

Standard deviation for the 24 countries is 23.83

f)

Standard deviation

5 0

3 years ago