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3 years ago

Write an inequality for x that would give this rectangle a perimeter of at least 300ft

Mathematics

1 answer:

Travka [436]3 years ago

7 0

Answer:

The inequality for $x$ is:

$x\geq 142$

Step-by-step explanation:

Given:

Width of rectangle = 3 ft

Height or length of rectangle = $(x+5)$ ft

Perimeter is at least 300 ft

To write an inequality for $x$ .

Solution:

Perimeter of a rectangle is given as:

⇒ $2l+2w$

where $l$ represents length of the rectangle and $w$ represents the width of the rectangle.

Plugging in the given values in the formula, the perimeter can be given as:

⇒ $2(x+5)+2(3)$

Using distribution:

⇒ $2x+10+6$

Simplifying.

⇒ $2x+16$

The perimeter is at lest 300 ft. So, the inequality can be given as:

⇒ $2x+16\geq 300$

Solving for $x$

Subtracting both sides by 16.

⇒ $2x+16-16\geq 300-16$

⇒ $2x\geq284$

Dividing both sides by 2.

⇒ $\frac{2x}{2}\geq \frac{284}{2}$

⇒ $x\geq 142$ (Answer)

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Answer:

91.8 feet^3

Step-by-step explanation:

We can find out the area of a circle and then multiple that by the height to get the volume of the tree trunk (which is a cylinder).

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3.14 x 1.5^2 = 7.065

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If I get paid $8.00 per hour and I also get paid time-and-a-half for overtime hours, how much would I get paid for each overtime

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3 years ago

The diameters of bolts produced in a machine shop are normally distributed with a mean of 5.65 millimeters and a standard deviat

nexus9112 [7]

Answer:

Two diameters that separate the top 4% and the bottom 4% are 5.77 and 5.53 respectively.

Step-by-step explanation:

We are given that the diameters of bolts produced in a machine shop are normally distributed with a mean of 5.65 millimeters and a standard deviation of 0.07 millimeters.

Let X = diameters of bolts produced in a machine shop

So, X ~ N( $\mu=5.65,\sigma^{2} = 0.07^{2}$ )

The z score probability distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean

$\sigma$ = standard deviation

Now, we have to find the two diameters that separate the top 4% and the bottom 4%.

Firstly, Probability that the diameter separate the top 4% is given by;

P(X > x) = 0.04

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-5.65}{0.07}$ ) = 0.04

P(Z > $\frac{x-5.65}{0.07}$ ) = 0.04

So, the critical value of x in z table which separate the top 4% is given as 1.7507, which means;

$\frac{x-5.65}{0.07}$ = 1.7507

$x-5.65 = 0.07 \times 1.7507$

$x$ = 5.65 + 0.122549 = 5.77

Secondly, Probability that the diameter separate the bottom 4% is given by;

P(X < x) = 0.04

P( $\frac{X-\mu}{\sigma}$ < $\frac{x-5.65}{0.07}$ ) = 0.04

P(Z < $\frac{x-5.65}{0.07}$ ) = 0.04

So, the critical value of x in z table which separate the bottom 4% is given as -1.7507, which means;

$\frac{x-5.65}{0.07}$ = -1.7507

$x-5.65 = 0.07 \times (-1.7507)$

$x$ = 5.65 - 0.122549 = 5.53

Therefore, the two diameters that separate the top 4% and the bottom 4% are 5.77 and 5.53 respectively.

4 0

3 years ago

The CEO of a large electric utility claims that more than 75 percent of his customers are very satisfied with the service they r

vichka [17]

Answer:

$z=\frac{0.77-0.75}{\sqrt{\frac{0.75(1-0.75)}{100}}}=0.462$

Now we can find the p value using the alternative hypothesis with this probability:

$p_v =P(z>0.462)=0.322$

Since the p value is large enough, we have evidence to conclude that the true proportion for this case is NOT significanctly higher than 0.75 since we FAIL to reject the null hypothesis at any significance level lower than 30%

Step-by-step explanation:

Information provided

n=100 represent the random sample selected

$\hat p=0.77$ estimated proportion of students that are satisfied

is the value that we want to test

z would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to verify if more than 75 percent of his customers are very satisfied with the service they receive, then the system of hypothesis is.:

Null hypothesis: $p\leq 0.75$

Alternative hypothesis: $p > 0.75$

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing the info given we got:

$z=\frac{0.77-0.75}{\sqrt{\frac{0.75(1-0.75)}{100}}}=0.462$

Now we can find the p value using the alternative hypothesis with this probability:

$p_v =P(z>0.462)=0.322$

Since the p value is large enough we have evidence to conclude that the true proportion for this case is NOT significanctly higher than 0.75 since we FAIL to reject the null hypothesis at any significance level lower than 30%

4 0

3 years ago