Pablo shares 525 cupcakes in the ratio of 3:4 between his

Emily's family loves to work together in the garden. They have a slight preference for flowers, as 60percent of their plants are

DaniilM [7]

Answer:

There are 20 vegetable plants in garden.

Step-by-step explanation:

We are given the following in the question:

Percentage of flowers = 60%

Percentage of vegetable = 40%

Number of plants in garden = 50

Number of vegetables in garden =

$40\% \times \text{Number of plant}\\\\= \dfrac{40}{100}\times 50\\\\= 20$

Number of flowers in garden =

$60\% \times \text{Number of plant}\\\\= \dfrac{60}{100}\times 50\\\\= 30$

Thus, there are 20 vegetable plants in garden.

6 0

2 years ago

The regular octagon has a perimeter of 122.4 cm. A regular octagon with a radius of 20 centimeters and a perimeter of 122.4 cent

Mamont248 [21]

A C and E

Step-by-step explanation:

I took the test

6 0

2 years ago

Read 2 more answers

Boris chooses three DIFFERENT numbers.

g100num [7]

Answer:

4, 5, 27.

Step-by-step explanation:

Let us find our numbers using given information.

Factors of 20 are: 1,2,4,5,10 and 20.

Let us try by taking number 4 and 5.

$36-(4+5)=36-9=27$

We can see that $3^{3} =27$

Now let us find sum of these numbers.

$4+5+27=36$

Therefore, numbers chosen by Boris will be 4, 5 and 27 as these numbers satisfies our all conditions.

6 0

2 years ago

In a certain classroom ,12.5% of the students own at least one pet. If there are 32 students in the classroom,how many students

Y_Kistochka [10]

Answer:

28

Step-by-step explanation:

Given that 12.5% of students own at least 1 pet, then the percentage of students who do not own a pet is

100% - 12.5% = 87.5%

Calculate 87.5% of 32, that is

$\frac{87.5}{100}$ × 32

= 0.875 × 32 = 28 ← number of students who do not own a pet

8 0

2 years ago

Read 2 more answers

A random sample of 35 undergraduate students who completed two years of college were asked to take a basic mathematics test. The

lisov135 [29]

Answer:

$(75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96$

$(75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96$

And the confidence interval would be given by:

$-1.96 \leq \mu_1 -\mu_2 \leq 7.96$

And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.

Step-by-step explanation:

For this case we have the following info given :

$\bar X_1= 75.1$ represent the sample mean for the scores of the undergraduate students

$s_1 = 12.8$ represent the standard deviation for the undergraduate students

$n_1 =35$ the sample size for the undergraduate

$\bar X_2= 72.1$ represent the sample mean for the scores of the high school students

$s_2 = 14.6$ represent the standard deviation for the high school students

$n_2 =50$ the sample size for the high school

The confidence interval for the true difference of means is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

The degrees of freedom are given by:

$df=n_1 +n_2 -2= 35+50-2=83$

The confidence level is 90% and the significance level is $\alpha=0.1$ and $\alpha/2 =0.05$ then the critical value would be:

$t_{\alpha/2}= 1.99$

And replacing the info we got:

$(75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96$

$(75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96$

And the confidence interval would be given by:

$-1.96 \leq \mu_1 -\mu_2 \leq 7.96$

And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.

8 0

2 years ago