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3 years ago

Can someone help like i posted this 5x and no one is helping like yall all around the world help mehh!!

Mathematics

2 answers:

liq [111]3 years ago

8 0

Answer:

well I would see what a average tud holds and then multiple till you get the total amount or round up to

Tresset [83]3 years ago

6 0

Answer:

It is impossible to answer this question without knowing the capacity of the tub. If I have this info I can easily solve the problem.

Step-by-step explanation:

the equation would be

9.5/90 = (capacity)/x where x= the amount of seconds it takes to fill the tub.

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Find and interpret a 95% confidence interval to estimate the average number of bolts per box for all boxes in the population. Ro

gayaneshka [121]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The 95% confidence interval is $49.85 < \mu < 54.15$

This means that there is 95% chance that the true population mean is within this interval

Step-by-step explanation:

From the question we are told that

The sample size is n = 30

The sample mean is $\= x = 52$

The population standard deviation is $\sigma = 6$

Given that the confidence level is 95% then the level of confidence is evaluate as

$\alpha = 100 - 95$

$\alpha = 5\%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table , the values is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

substituting values

$E = 1.96 * \frac{ 6 }{\sqrt{30} }$

$E = 2.147$

The 95% confidence interval is mathematically represented as

$\= x - E < \mu < \= x + E$

substituting values

$52 - 2.147 < \mu < 52 + 2.147$

$49.85 < \mu < 54.15$

4 0

4 years ago

Find the approximate side length of a square game board with an area of 186in2

Arturiano [62]

13.6in is the answer

8 0

4 years ago

Read 2 more answers

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Komok [63]

Y = 229 people x = hours after

7 0

4 years ago

Read 2 more answers

At a local sandwich shop the average customer spends $25 at dinner with a standard deviation of $2.30. In a simple random sample

oee [108]

Answer:

The probability that the mean receipt for dinner is between $24.50 and $25.50 is

P(24.50≤ x ≤ 25.50)= 0.8324

Step-by-step explanation:

Step(i):-

Given mean of Population(μ) = $25

Given standard deviation of the Population (σ) = 2.30

Sample size 'n' = 40

Let 'X' be the random variable in normal distribution

Given X₁ = 24.50

$Z_{1} =\frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{24.50-25}{\frac{2.30}{\sqrt{40} } } = -1.375$

Given X₂ = 25.50

$Z_{2} =\frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{25.50-25}{\frac{2.30}{\sqrt{40} } } = 1.3751$

Step(ii):-

The probability that the mean receipt for dinner is between $24.50 and $25.50

P(x₁≤ x ≤ x₂) = P(Z₁≤ z ≤ Z₂) = A(Z₂) + A(z₁)

P(24.50≤ x ≤ 25.50) = P(-1.375≤ z ≤ 1.375) = A(1.375) + A(1.375)

P(24.50≤ x ≤ 25.50)= 2 A( 1.375)

= 2 × 0.4162 ( from normal table)

= 0.8324

Final answer:-

The probability that the mean receipt for dinner is between $24.50 and $25.50 is 0.8324

4 0

3 years ago

The junior varsity football team score 23 points in last Saturday's game they score the combination of seven point touchdowns in

frutty [35]

Answer:

2 touchdowns, 3 field goals

Step-by-step explanation:

The number of touchdowns cannot be more than 3, so it is relatively easy to find the solution by trial and error.

23 is not divisible by 3, so 0 touchdowns is not a solution

23 -7 = 16 is not divisible by 3, so 1 touchdown is not a solution

23 -14 = 9 is divisible by 3, so 2 touchdowns and 3 field goals is a solution

21 -21 = 2 is not divisible by 3, so there is only one solution.

7 0

3 years ago