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30 subtracted from 4 times a number is greater than the opposite of 10

irakobra [83]

If you're trying to make an inequality, it would be:

30 - 4x > -10.

4 0

3 years ago

Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r

ZanzabumX [31]

Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean weight = $\frac{\sum X}{n}$ = 21.88 ounces

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 0.201 ounces

n = sample of boxes = 12

$\mu$ = population mean weight

Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.796 < $t_1_1$ < 1.796) = 0.90 {As the critical value of t at 11 degrees of

freedom are -1.796 & 1.796 with P = 5%}

P(-1.796 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.796) = 0.90

P( $-1.796 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $21.88-1.796 \times {\frac{0.201}{\sqrt{12} } }$ , $21.88+1.796 \times {\frac{0.201}{\sqrt{12} } }$ ]

= [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

8 0

3 years ago

What is 45.13 rounded to the nearest one?

SCORPION-xisa [38]

Answer:

45

Step-by-step explanation:

8 0

3 years ago

Harold, a marina manager, purchased two boats. He then sold the boats, the first

zaharov [31]

Harold paid $ 16,632 and $ 38,808 for each of the boats.

Since Harold, a marina manager, purchased two boats, and he then sold the boats, the first at a profit of 40% and the second at a profit of 60%, and the total profit on the sale of the two boats was 54 % and $ 88 704 was the total selling price of the two boat, to determine what did Harold originally pay for each of the two boats the following calculation must be performed:

55 x 0.6 + 45 x 0.4 = 51
65 x 0.6 + 35 x 0.4 = 53
70 x 0.6 + 35 x 0.4 = 54

88,704 x 0.7 = 62,092.80
160 = 62,092.80
100 = X
100 x 62,092.80 / 160 = X
38.808 = X

88,704 x 0.3 = 26,611.20
140 = 26,611.20
100 = X
100 x 26,611.20 / 160 = X
16,632 = X

Therefore, Harold paid $ 16,632 and $ 38,808 for each of the boats.

Learn more in brainly.com/question/21446500

7 0

3 years ago

Find the values of x and y X= Y=

Cerrena [4.2K]

Answer:

x = 7

y = 4

Step-by-step explanation:

y + 12 = 3x - 5

5y - 4 = 3x - 5

So y + 12 = 5y - x

To solve the equation above, y =4

Sub y = 4,

x = 7

7 0

3 years ago