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3 years ago

Lol I wasnt paying attention pleaseeee help

Mathematics

2 answers:

stiv31 [10]3 years ago

7 0

Answer:

3 1/3 pizzas

Step-by-step explanation:

Take 2/3 and multiply it by 5/1. You get 10/3, which is an improper fraction. You take that and simplify it. I know 9 goes into 3 3 times, so that means those boys had 3 1/3 pizzas.

(Can I get a brainliest?)

mr_godi [17]3 years ago

5 0

Answer:

3 1/3

Step-by-step explanation:

super easy, just multiply them

5/1*1/3 and you get 10/3 and you simplify and get 3 1/3 pizza

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Complete Question

The complete question is shown on the first and second uploaded image

Answer:

Part 1

The correct option is B

Part 11

The correct option is H

Step-by-step explanation:

From the question we are told that

The sample size is n = 15

Generally the sample mean for the input temperature is mathematically represented as

$\= x _1 = \frac{\sum x_i}{n}$

$\= x _1 = \frac{57.6 + 68.9 \cdots +60.4 }{15}$

$\= x _1 = 62.57$

Generally the sample mean for the output temperature is mathematically represented as

$\= x _2 = \frac{\sum x_i}{n}$

$\= x _2 = \frac{65.1 + 74.4 \cdots +67.3 }{15}$

$\= x _2 = 55.97$

Generally the difference between the mean of the input temperature and that of the output temperature is

$d = \= x_1 - \= x_2$

=> $d = 62.57 - 55.97$

=> $d = 6.6$

Generally the standard deviation of the difference between the input temperature and the output temperature is mathematically represented as

$s_d = \sqrt {\frac{1}{n-1 } \sum [d_i - d]^2}$

=> $s_d = \sqrt{\frac{[(57.6 - 65.1) - 6.6]^2+[(68.9 - 74.4) - 6.6]^2+ \cdots +[(68.1 - 74.7) - 6.6]^2 }{15-1} }$

=> $s_d = 1.732$

The null hypothesis is $H_o : \mu_1 -\mu_2 = 6$

The alternative hypothesis is $H_a : \mu _1 - \mu_2\ne 6$

Generally the test statistics is mathematically represented as

$t = \frac{d - 6}{ \frac{s_d}{ \sqrt{n} } }$

$t = \frac{6.6- 6}{ \frac{1.732}{ \sqrt{15} } }$

$t = 1.342$

Generally the p-value is mathematically represented as

$p-value = 2 * P(t > 1.342)$

From the student t distribution table( reference - danielsoper(dot)com(slash)statcalc(slash)calculator) at a degree of freedom of df = n-1 = 15-1 = 14

$P(t > 1.342) = t_{1.342 , 14} = 0.100478$