All categories

3 years ago

A zookeeper divided 2,440 pounds of food equally among 8 elephants. How many pounds of food did each elephant receive

Mathematics

2 answers:

spin [16.1K]3 years ago

7 0

305 pounds in total Because 2440 divided by 8 is 305

denis-greek [22]3 years ago

4 0

Each elephant received 305 pounds of food.

You might be interested in

What Is the Answer To number 1 ?

DochEvi [55]

Answer: 4 apples for $3.64

Step-by-step explanation:

The best buy is going to be the one that costs the least. To figure this out, you divide the number of apples by the price. (do not round)

4/3.64=1.09

9/8.01=1.23

14/12.43=1.12

24/21.12=1.13

As you can see, the 4 apples for $3.64 is the better choice since the number is the least.

5 0

3 years ago

Please help. 7. 2 × 5. 28

mr_godi [17]

Answer:

The Answer is 10! (I Don't know if you meant 2x5 or 2x5x28) If I am wrong please let me know! If you meant 2x5x28 the answer is 280!

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Factor completely w^4-2401

Verdich [7]

$w^4-2401=(w^2)^2-49^2=(w^2-49)(w^2+49)=(w^2-7^2)(w^2+49)\\\\=(w-7)(w+7)(w^2+49)\\\\------------------------------\\\\a^2-b^2=(a-b)(a+b)$

6 0

3 years ago

Read 2 more answers

What is the x intercept of y= 4x+8

ankoles [38]

To find the x-intercept what you do is you plug in zero for Y:
0=4x-8
Then you would add 8 to the left side and x =2

7 0

3 years ago

Many people think that a national lobby's successful fight against gun control legislation is reflecting the will of a minority

Andre45 [30]

Answer:

The 95% confidence interval of the true proportion of all Americans who are in favor of gun control legislation is (0.5471, 0.5779).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence interval $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

A random sample of 4000 citizens yielded 2250 who are in favor of gun control legislation. This means that $n = 4000$ and $\pi = \frac{2250}{4000} = 0.5625$

Estimate the true proportion of all Americans who are in favor of gun control legislation using a 95% confidence interval

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5625 - 1.96\sqrt{\frac{0.5625*0.4375}{4000}} = 0.5471$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5625 + 1.96\sqrt{\frac{0.5625*0.4375}{4000}} = 0.5779$

The 95% confidence interval of the true proportion of all Americans who are in favor of gun control legislation is (0.5471, 0.5779).

8 0

3 years ago