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3 years ago

If it takes 930 kg of food to feed a pair of elephants for 3 days how much food would u need to feed them for a week?

Mathematics

2 answers:

bagirrra123 [75]3 years ago

3 0

Answer:

217 kg

Step-by-step explanation:

i just know the answer

lapo4ka [179]3 years ago

3 0

2,170 kg because
3/930=310
310 is for one day so you multiply it by 7 to get a week
310*7=2,170

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2 years ago

If the width and length of a rectangle are both increased by 10%, by what percent does the area of the rectangle increase?

S_A_V [24]

Answer:

10%

Step-by-step explanation:

We know Perimeter of rectangle

= 2 (Length + Width)

If width and length are increased by 10%

New length= L + 10/100L = (1.1) L

New width = W + 10/100W = (1.1) W

Perimeter = 2 [(1.1) L + (1.1) W]

Perimeter = 2 (1.1) (L + W)

Perimeter = (2.2) (L + W)

Increase in perimeter = 2.2 (L + W) - 2 (L + W)

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2 years ago

1. Which of the following equations has a solution of x = -5? Equation A------> 16x-7=11x-32 Equation B------> -4x- 10=2x+

Nitella [24]

Answer:

Step-by-step explanation:

Equation A

16x - 7 = 11x - 32

(16x - 7) - 11x = (11x - 32) -11x

5x - 7 = -32

(5x - 7) + 7 = (-32) + 7

5x = -25

(5x)/5 = (-25)/5

x = -5

Equation B

-4x - 10 = 2x + 20

(-4x - 10) - 2x = (2x + 20) - 2x

-6x - 10 = 20

(-6x - 10) + 10 = (20) + 10

-6x = 30

(-6x)/-6 = (30)/-6

x = -5

Both Equation A and Equation B have a solution of -5.

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3 years ago

Name the property the equation illustrates. 8.2 + (–8.2) = 0

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3 years ago

Read 2 more answers

Two statistics teachers both believe that each has the smarter class. To put this to the test, they give the same final exam to

SVETLANKA909090 [29]

Answer:

We conclude that there is no difference between the two classes.

Step-by-step explanation:

We are given that two statistics teachers both believe that each has a smarter class.

A summary of the class sizes, class means, and standard deviations is given below:n1 = 47, x-bar1 = 84.4, s1 = 18n2 = 50, x-bar2 = 82.9, s2 = 17

Let $\mu_1$ = mean age of student cars.

$\mu_2$ = mean age of faculty cars.

So, Null Hypothesis, $H_0$ : $\mu_1=\mu_2$ {means that there is no difference in the two classes}

Alternate Hypothesis, $H_A$ : $\mu_1\neq \mu_2$ {means that there is a difference in the two classes}

The test statistics that will be used here is Two-sample t-test statistics because we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean age of student cars = 8 years

$\bar X_2$ = sample mean age of faculty cars = 5.3 years

$s_1$ = sample standard deviation of student cars = 3.6 years

$s_2$ = sample standard deviation of student cars = 3.7 years

$n_1$ = sample of student cars = 110

$n_2$ = sample of faculty cars = 75

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(47-1)\times 18^{2}+(50-1)\times 17^{2} }{47+50-2} }$ = 17.491

So, the test statistics = $\frac{(84.4-82.9)-(0)}{17.491 \times \sqrt{\frac{1}{47}+\frac{1}{50} } }$ ~ $t_9_5$

= 0.422

The value of t-test statistics is 0.422.

Now, the P-value of the test statistics is given by;

P-value = P( $t_9_5$ > 0.422) = From the t table it is clear that the P-value will lie somewhere between 40% and 30%.

Since the P-value of our test statistics is way more than the level of significance of 0.04, so we have insufficient evidence to reject our null hypothesis as our test statistics will not fall in the rejection region.

Therefore, we conclude that there is no difference between the two classes.

6 0

2 years ago