Ask question

Ask question

All categories

antoniya [11.8K]

3 years ago

13

Five friends plan to split a restaurant bill evenly. The total cost of the meal is $89.75, and they want to leave a 20% tip. Wha

t amount should each friend pay? EXPLAIN

2 answers:

sladkih [1.3K]3 years ago

8 0

The answer is 21.54

Hope this helped

Vikki [24]3 years ago

8 0

$21.54

$89.75+17.95 (20% times 89.75) = $107.70 . Divide this by 5

You might be interested in

Suppose 40% of all college students have a computer at home and a sample of 100 is taken. what is the probability that more than

scoundrel [369]

unlikely roughtly 40% have computers you cant make up for that other 10 percent

4 0

3 years ago

Please help me with this homework

Nady [450]

Answer:

Rational

Step-by-step explanation:

The number comes to a stop (doesn't keep going), so it would be rational.

7 0

3 years ago

Read 2 more answers

The question is in the picture above. Please help

kow [346]

Answer:

it is 1

Step-by-step explanation:

5 0

3 years ago

Please help this is including substitution and system of equations.

Darina [25.2K]

Answer: I gotcha

Step-by-step explanation:

B, C, C, D, B

I just helped my brother with this

6 0

3 years ago

A national grocery store chain wants to test the difference in the average weight of turkeys sold in Detroit and the average wei

Arte-miy333 [17]

Answer:

<em>Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom</em>

<em>The null hypothesis is accepted . </em>

<em>Assume the population variances are approximately the same</em>

<u><em>Step-by-step explanation:</em></u>

<u>Explanation</u>:-

Given data a random sample of 20 turkeys sold at the chain's stores in Detroit yielded a sample mean of 17.53 pounds, with a sample standard deviation of 3.2 pounds

<em>The first sample size 'n₁'= 20</em>

<em>mean of the first sample 'x₁⁻'= 17.53 pounds</em>

<em>standard deviation of first sample S₁ = 3.2 pounds</em>

Given data a random sample of 24 turkeys sold at the chain's stores in Charlotte yielded a sample mean of 14.89 pounds, with a sample standard deviation of 2.7 pounds

<em>The second sample size n₂ = 24</em>

<em>mean of the second sample "x₂⁻"= 14.89 pounds</em>

<em>standard deviation of second sample S₂ = 2.7 poun</em>ds

<u><em>Null hypothesis</em></u><u>:-</u><u><em>H₀</em></u><em>: The Population Variance are approximately same</em>

<u><em>Alternatively hypothesis</em></u><em>: H₁:The Population Variance are approximately same</em>

<em>Level of significance ∝ =0.05</em>

<em>Degrees of freedom ν = n₁ +n₂ -2 =20+24-2 = 42</em>

<em>Test statistic :-</em>

<em> </em> $t = \frac{x^{-} _{1} - x_{2} }{\sqrt{S^2(\frac{1}{n_{1} } }+\frac{1}{n_{2} } }$

<em> where </em> $S^{2} = \frac{n_{1} S_{1} ^{2}+n_{2}S_{2} ^{2} }{n_{1} +n_{2} -2}$

$S^{2} = \frac{20X(3.2)^2+24X(2.7)^2}{20+24-2}$

<em> substitute values and we get S² = 40.988</em>

<em> </em> $t= \frac{17.53-14.89 }{\sqrt{40.988(\frac{1}{20} }+\frac{1}{24} )}$ <em></em>

<em> </em> t = 1.3622

Calculated value t = 1.3622

Tabulated value 't' = 2.081

Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom

<u><em>Conclusion</em></u>:-

<em>The null hypothesis is accepted </em>

<em>Assume the population variances are approximately the same.</em>

<em> </em>

<em> </em>

<em> </em>

6 0

3 years ago