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3 years ago

Do vegetarians eat animal crackers?

Mathematics

2 answers:

mr Goodwill [35]3 years ago

8 0

Answer:

Good question. Maybe.

Step-by-step explanation:

Maybe they eat in secret.

Just Don't tell them it's an animal and they will eat it.

vovikov84 [41]3 years ago

6 0

Answer:

yes

Step-by-step explanation:

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One pound of tomatoes costs $1.59.One pound of bananas costs $1.87. How much more do 5 pounds of tomatoes costs than 2 pounds of

Bad White [126]

when you multiply $1.59 by 5 you get $7.95

when you multiply $1.87 by 2 you get $4.74

so when you subtract them you get $3.21

3 0

3 years ago

24x3 + 18x2 - 6x Factor Polynomials Completely

andreev551 [17]

You answer for 24x3+18x2-6x would be 102x

8 0

3 years ago

Kidney beans costs $1.13 per kilogram. What equation shows the relationship between total cost (t), price (p), and kilograms pur

sukhopar [10]

Total cost = price per kilogram• how many kilograms were purcest
t= p•k
t= 1.13•k

7 0

4 years ago

The length of the box is 2x - 2 and the width is x - 5. Find the perimeter. (HINT: p= 2L + 2W) Choose one.

trasher [3.6K]

Answer:

Option D: 6x-14 is the correct answer.

Step-by-step explanation:

Given that:

Length of the box = L = 2x-2

Width of the box = W = x-5

Perimeter is defined by the sum of length of all sides of the box.

Perimeter of the box = P = 2L + 2W

Perimeter of box = 2(2x-2) + 2(x-5)

Perimeter = 4x-4+2x-10

Perimeter = 6x-14

The perimeter of the box is 6x-14.

Hence,

Option D: 6x-14 is the correct answer.

6 0

3 years ago

Randomly selected 110 student cars have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly select

monitta

Answer:

1. Yes, there is sufficient evidence to support the claim that student cars are older than faculty cars.

2. The 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].

Step-by-step explanation:

We are given that randomly selected 110 student cars to have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly selected 75 faculty cars to have ages with a mean of 5.3 years and a standard deviation of 3.7 years.

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

$\mu_2$ = mean age of faculty cars.

So, Null Hypothesis, $H_0$ : $\mu_1 \leq \mu_2$ {means that the student cars are younger than or equal to faculty cars}

Alternate Hypothesis, $H_A$ : $\mu_1>\mu_2$ {means that the student cars are older than faculty cars}

(1) 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, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(110-1)\times 3.6^{2}+(75-1)\times 3.7^{2} }{110+75-2} }$ = 3.641

So, the test statistics = $\frac{(8-5.3)-(0)} {3.641 \times \sqrt{\frac{1}{110}+\frac{1}{75} } }$ ~ $t_1_8_3$

= 4.952

The value of t-test statistics is 4.952.

Since the value of our test statistics is more than the critical value of t, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we support the claim that student cars are older than faculty cars.

(2) The 98% confidence interval for the difference between the two population means ( $\mu_1-\mu_2$ ) is given by;

98% C.I. for ( $\mu_1-\mu_2$ ) = $(\bar X_1-\bar X_2) \pm (t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} })$

= $(8-5.3) \pm (2.326 \times 3.641 \times \sqrt{\frac{1}{110}+\frac{1}{75} })$

= $[2.7 \pm 1.268]$

= [1.432, 3.968]

Here, the critical value of t at a 1% level of significance is 2.326.

Hence, the 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].

7 0

3 years ago