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3 years ago

Helping others makes a person good

Mathematics

1 answer:

oksano4ka [1.4K]3 years ago

3 0

Answer:

9 cm.

Step-by-step explanation:

Let the height of the pyramid be h cm, then the height of the cuboid is (15 - h) cm.

Volume of the pyramid:

= 1/3 * h * s^2 where s is the length of a side of the square base.

= hs^2/3 cm^3

Volume of the cuboid:

= s^2(15 - h).

So we have:

hs^2/ 3 = 300.......................(1)

s^2(15 - h) = 600..................(2)

From equation (1) :

h s^2 = 900

s^2 = 900/h

Now substitute for s^2 in equation (2) :

(900/h)(15 - h) = 600

Multiply through by h:

900(15 - h) = 600h

13500 - 900h = 600h

1500h = 13500

h = 9 cm (answer). brainliest?

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Select the expressions that are equivalent to 18m - 12.

krok68 [10]

Answer:

All the expressions other than option E, is equivalent to the expression 18m - 12.

Step-by-step explanation:

A. 6m - 4 + 6m -4 + 6m - 4

or 6m+6m+6m -4 -4 -4

or 18m -12

B. 12m + 6 - 6m -6

or 12m - 6m + 6 - 6

or 6m

C.6(3m - 2)

or 18m - 12

D.3(6m - 4)

or 18m - 12

E. 24n - 4² + 8 -6m

This option can not satisfy the given expression as it contains another variable as n.

4 0

3 years ago

A driver's education course compared 1,500 students who had not taken the course with 1,850 students who had. Of those students

aivan3 [116]

Answer:

P-value is less than 0.999995 .

Step-by-step explanation:

We are given that a driver's education course compared 1,500 students who had not taken the course with 1,850 students who had.

Null Hypothesis, $H_0$ : $p_1 = p_2$ {means students who took the driver's education course and those who didn't took have same chances to pass the written driver's exam the first time}

Alternate Hypothesis, $H_1$ : $p_1 > p_2$ {means students who took the driver's education course were more likely to pass the written driver's exam the first time}

The test statistics used here will be two sample Binomial statistics i.e.;

$\frac{(\hat p_1 - \hat p_2)-(p_1 - p_2)}{\sqrt{\frac{\hat p_1(1- \hat p_1)}{n_1} + \frac{\hat p_2(1- \hat p_2)}{n_2} } }$ ~ N(0,1)

Here, $\hat p_1$ = No. of students passed the exam ÷ No.of Students that had taken the course

$\hat p_1$ = $\frac{1150}{1850}$ Similarly, $\hat p_2$ = $\frac{1440}{1500}$ $n_1$ = 1,850 $n_2$ = 1,500

Test Statistics = $\frac{(\frac{1150}{1850} -\frac{1440}{1500})-0}{\sqrt{\frac{\frac{1150}{1850}(1- \frac{1150}{1850})}{1850} + \frac{\frac{1440}{1500}(1- \frac{1440}{1500})}{1500} } }$ = -27.38

P-value is given by, P(Z > -27.38) = 1 - P(Z > 27.38)

Now, in z table the highest critical value given is 4.4172 which corresponds to the probability value of 0.0005%. Since our test statistics is way higher than this so we can only say that the p-value for an appropriate hypothesis test is less than 0.999995 .

6 0

3 years ago

Solve for a:<br>9+2a = -3-4a

lubasha [3.4K]

Answer:

a=-2

Step-by-step explanation:

Let me know if you need the steps tho.

5 0

3 years ago

Which of the following is the graph of the quadratic function? y=x^2+4-12

tiny-mole [99]

Answer:

Graph B

Step-by-step explanation:

The solutions to a quadratic equation are the points on which we have the graph of the curve touching the x-axis

Now the first thing we will do here is to solve the quadratic equation graph;

x^2 + 4x -12 = 0

x^2 +6x - 2x -12 = 0

x (x + 6) -2(x + 6) = 0

(x-2)(x + 6) = 0

x = 2 or -6

So the graph that touches the x-axis at the points x = 2 and x = -6 is the solution to the quadratic equation

Graph B is the closest to what we have as answer

5 0

3 years ago

In 2001 , chipper had 102 runs batted in. In 2008, Chipper had only 75 runs batted in. Calculate the percent change in runs batt

DerKrebs [107]

Answer:

The percent change in runs batted in from 2001 to 2008 is 26.47%

Step-by-step explanation:

The number of runs chipper batted in the year 2001 = 102 runs

The number of runs chipper batted in the year 2008 = 75 runs

Here, so the Change in the runs batted

= Runs Batted in year 2001 - Runs batted in the year 2008

= 102 runs - 75 runs = 27 runs

Now, The percentage change in the run = $= \frac{\textrm{Change in the batted run from 2001 to 2008}}{\textrm{Total number of batted runs in 2001}} \times 100$

= $\frac{27}{102} \times 100 = 26.47$

or, The percentage change = 26.47%

Hence, the percent change in runs batted in from 2001 to 2008 is 26.47%

5 0

3 years ago

Helping others makes a person good ​

Helping others makes a person good