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yan [13]
3 years ago
8

A survey of all beings on the planet mimstoon found that 10 beings preferred lirt juice to all other juices If 25 beings are sur

veyed altogether what percent of them preferred Lirt juice
Mathematics
1 answer:
kherson [118]3 years ago
4 0

Answer:

The percentage of the surveyed that preferred Lirt juice is 40%

Step-by-step explanation:

In this question, we are asked to calculate the percentage of beings that preferred Lirt juice

Let’s look at the survey presented in the question. 10 beings preferred out of a total of 25

The percentage that preferred Lirt juice will thus be;

10/25 * 100%

= 10 * 4 = 40%

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2x+5=11 two step equations<br>​
Sati [7]

Answer:

x=3

Explanation: 11-5=6

6/2= 3

Plug 3 into the equation and you get 2x3+5=11 which 2x3=6 and 6+5=11

4 0
3 years ago
I'd love some help on this question. I do not understand how you would go about working this out, so please explain the working
zhannawk [14.2K]
Area = length x width and volume = length x width x height soo you would have to multiple 4 and 25 and that’s 100
7 0
3 years ago
Connecticut families were asked how much they spent weekly on groceries. Using the following data, construct and interpret a 95%
Amanda [17]

Answer:

The 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

Step-by-step explanation:

The data for the amount of money spent weekly on groceries is as follows:

S = {210, 23, 350, 112, 27, 175, 275, 50, 95, 450}

<em>n</em> = 10

Compute the sample mean and sample standard deviation:

\bar x =\frac{1}{n}\cdot\sum X=\frac{ 1767 }{ 10 }= 176.7

s= \sqrt{ \frac{ \sum{\left(x_i - \overline{x}\right)^2 }}{n-1} }       = \sqrt{ \frac{ 188448.1 }{ 10 - 1} } \approx 144.702

It is assumed that the data come from a normal distribution.

Since the population standard deviation is not known, use a <em>t</em> confidence interval.

The critical value of <em>t</em> for 95% confidence level and degrees of freedom = n - 1 = 10 - 1 = 9 is:

t_{\alpha/2, (n-1)}=t_{0.05/2, (10-1)}=t_{0.025, 9}=2.262

*Use a <em>t</em>-table.

Compute the 95% confidence interval for the population mean amount spent on groceries by Connecticut families as follows:

CI=\bar x\pm t_{\alpha/2, (n-1)}\cdot\ \frac{s}{\sqrt{n}}

     =176.7\pm 2.262\cdot\ \frac{144.702}{\sqrt{10}}\\\\=176.7\pm 103.5064\\\\=(73.1936, 280.2064)\\\\\approx (73.20, 280.21)

Thus, the 95% confidence interval for the population mean amount spent on groceries by Connecticut families is ($73.20, $280.21).

7 0
3 years ago
If 4 raffle tickets is 6 dollars how much is 1 raffle ticket
sertanlavr [38]
4/6 = 1/x cross multiple so x= 1.5 dollars
8 0
3 years ago
I am stuck on this question do you mind checking it for me
nataly862011 [7]

Solution:

Given:

\frac{a}{b}\cdot\frac{c}{d}

Since b and d are nonzero elements, then it is the product of two rational numbers.

Multiplying two rational numbers produces another rational number.

Therefore, the product is a rational expression.

OPTION C is the correct answer.

4 0
1 year ago
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