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3 years ago

ILL GIVE BRAINLIST IF U ANSWER THIS QUESTION

Mathematics

1 answer:

Yakvenalex [24]3 years ago

7 0

57 degrees because 180-123= 57 degrees

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A company that packages salted and unsalted mixed nuts received a complaint that claimed that the company’s salted packages cont

Eduardwww [97]

Answer:

$H_0 : \mu_1 - \mu_2 = 0$

$H_a : \mu_1 - \mu_2 \ne 0$

Step-by-step explanation:

Let the salted package be represented with 1 and the unsalted, 2.

So:

$\mu_1:$ Mean of salted package

$\mu_2:$ Mean of unsalted package

Considering the given options, the null hypothesis is that which contains =.

So, the null hypothesis is:

$H_0 : \mu_1 - \mu_2 = 0$

The opposite of = is $\ne$ . So, the alternate hypothesis, is that which contains $\ne$

So, the alternate hypothesis is:

$H_a : \mu_1 - \mu_2 \ne 0$

4 0

3 years ago

Work out an estimate for the value of <br> 48.7 × 61.2/11.3

avanturin [10]

Answer:

$263.755$

Step-by-step explanation:

$48.7 \times 61.2 \div 11.3 \\ 2,980.44 \div 11.3 \\ = 263.755752$

hope this helps

brainliest appreciated

good luck! have a nice day!

4 0

3 years ago

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A restaurant has a special whereby both parents can eat for $20 and each child can eat for $5. Assuming a family group consists

nordsb [41]

4groups of adults and 7 kids

5 0

3 years ago

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What is the least common multiple of 3 and 2?

vovikov84 [41]

Answer:

The answer is 6

Step-by-step explanation:

Hope this helps

6 0

3 years ago

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Rita is selecting a sandwich at the deli. The deli has four types of meat, three types of cheese and two types of bread. A delux

zalisa [80]

Answer:

24 possible dishes

Step-by-step explanation:

Given

Total

$Meat = 4$

$Cheese = 3$

$Bread = 2$

Selection

$Meat = 1$

$Cheese =2$

$Bread = 1$

Required

Determine the number of possible combo

Rita wants to select 1 of 4 meat types,

So, the number of selections is: $^4C_1$

Rita wants to select 2 of 3 cheese types,

So, the number of selections is: $^3C_2$

Rita wants to select 1 of 2 cheese types,

So, the number of selections is: $^2C_1$

The total number of selections is:

$Total = ^4C_1 * ^3C_2 * ^2C_1$

Using

$^nC_r = \frac{n!}{(n-r)!r!}$

We have:

$Total = \frac{4!}{(4-1)!1!} * \frac{3!}{(3-2)!2!} * \frac{2!}{(2-1)!1!}$

$Total = \frac{4!}{3!1!} * \frac{3!}{1!2!} * \frac{2!}{(1!1!}$

$Total = \frac{4*3!}{3!*1} * \frac{3*2!}{1*2!} * \frac{2*1}{1*1}$

$Total = 4* 3 * 2$

$Total = 24$

<em>Hence, there are 24 possible dishes</em>

4 0

3 years ago

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