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4 years ago

Please answer this correctly

Mathematics

2 answers:

Kitty [74]4 years ago

8 0

Answer:

153 ft^2

Step-by-step explanation:

If the top side were 24 ft long and the right side 10 ft long, you'd have a yellow rectangle with area 10 ft * 24 ft = 240 ft^2.

Now we subtract the parts that are missing.

3 ft * 13 ft = 39 ft^2

3 ft * 16 ft = 48 ft^2

240 ft^2 - 39 ft^2 - 48 ft^2 = 153 ft^2

Pavlova-9 [17]4 years ago

5 0

Answer:

153 square feet is the area

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What is the missing value in x2y8 · x3y? = x5y5?

sweet-ann [11.9K]

Answer:

-3

Step-by-step explanation:

To find:

The value of '?' in the following expression:

$x^2y^8.x^3y^? = x^5y^5$

Solution:

First of all, let us have a look at a few rules of the exponents:

$1.\ a^b.a^c=a^{b+c}$

$2.\ a^b.a^{-c}=a^{b-c}$

Let us apply these rules and try to find out the missing value ('?') in the given expression.

Left Hand Side (LHS):

$x^2y^8.x^3y^?\\\Rightarrow x^2.x^3.y^8.y^?\\\Rightarrow x^{2+3}y^{8+?}\\\Rightarrow x^{5}y^{8+?}$

Now, let us compare the LHS with RHS, we get:

$x^{5}y^{8+?} = x^{5}y^{5}$

Dividing with $x^5$ on both sides:

$y^{8+?} = y^5\\\Rightarrow 8+? = 5\\\Rightarrow ? = 5-8 \\\Rightarrow ? =-3$

Therefore, the missing value in the expression is -3.

4 0

3 years ago

By law, an industrial plant can discharge not more than 500 gallons of waste water per hour, on the average, into a neighboring

Alisiya [41]

Answer:

We conclude that not more than 500 gallons of wastewater per hour, on average, is discharged into a neighboring lake.

Step-by-step explanation:

We are given that an industrial plant can discharge not more than 500 gallons of wastewater per hour, on average, into a neighboring lake.

Four one-hour periods are selected randomly over a period of one week. The following are observed:

1384, 683, 1534, 405

Let $\mu$ = population average gallons of wastewater discharged per hour

So, Null Hypothesis, $H_0$ : $\mu \leq$ 500 gallons {means that not more than 500 gallons of wastewater per hour, on the average, is discharged into a neighboring lake}

Alternate Hypothesis, $H_A$ : $\mu$ > 500 gallons {means that more than 500 gallons of wastewater per hour, on the average, is discharged into a neighboring lake}

The test statistics that will be used here is One-sample t-test statistics because we don't know about population standard deviation;

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

where, $\bar X$ = sample mean = $\frac{\sum X}{n}$ = 1001.5 gallons

s = sample standard deviation = $\sqrt{\frac{\sum (X - \bar X)^{2} }{n-1} }$ = 543.79

n = sample of periods = 4

So, the test statistics = $\frac{1001.5-500}{\frac{543.79}{\sqrt{4} } }$ ~ $t_3$

= 1.844

The value of t-test statistics is 1.844.

Since in the question we are not given with the level of significance so we assume it to be 5%. Now, at 0.05 level of significance, the t table gives a critical value of 2.353 at 3 degrees of freedom for the right-tailed test.

Since the value of our test statistics is less than the critical value of z as 1.844 < 2.353, so we have insufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that not more than 500 gallons of wastewater per hour, on average, is discharged into a neighboring lake.

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