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Helen
3 days ago
15

Line AB contains points A(4, 5) and B(9, 7). What is the slope of ? –negative StartFraction 5 Over 2 EndFraction –negative Start

Fraction 2 Over 5 EndFraction StartFraction 2 Over 5 EndFraction StartFraction 5 Over 2 EndFraction
Mathematics
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The area of ABED is 49 square units. Given AGequals9 units and ACequals10 ​units, what fraction of the area of ACIG is represent
Svet_ta [12734]

Answer:

The shaped region accounts for 7/18 of the area of ACIG.

Step-by-step explanation:

Refer to the attached diagram for further clarity on the problem.

Step 1

Determine the length of one side of square ABED.

We know that

AB=BE=ED=AD

The area of a square can be calculated as

A=b^{2}

where b is the side length.

We have

A=49\ units^2

So we substitute

49=b^{2}

b=7\ units

Thus,

AB=BE=ED=AD=7\ units

Step 2

Calculate the area of ACIG.

The area of rectangle ACIG is determined by

A=(AC)(AG)

Substituting the given values yields

A=(9)(10)=90\ units^2

Step 3

Determine the area of the shaded rectangle DEHG.

The area of rectangle DEHG is given by

A=(DE)(DG)

We find DE=7\ units

DG=AG-AD=9-7=2\ units

and substitute A=(7)(2)=14\ units^2

Step 4

Calculate the area of shaded rectangle BCFE.

The area of rectangle BCFE equals

A=(EF)(CF)

We see that

EF=AC-AB=10-7=3\ units

CF=BE=7\ units

and substitute

A=(3)(7)=21\ units^2

Step 5

Add the areas of the shaded regions together.

14+21=35\ units^2

Step 6

Divide the area of the shaded region by the area of ACIG.

\frac{35}{90}

Simplify this fraction by dividing both the numerator and denominator by 5.

\frac{7}{18}

Hence, the shaped region represents 7/18 of the area of ACIG.

5 0
2 months ago
Statistics can help decide the authorship of literary works. Sonnets by a certain Elizabethan poet are known to contain an avera
AnnZ [12381]

Response:

It is inferred that the authors of the sonnets belong to a certain poet from the Elizabethan era.

Step-by-step breakdown:

The details provided in the question are as follows:

Population mean, μ = 8.9

Sample mean, \bar{x} = 10.2

Sample size, n = 6

Alpha, α = 0.05

Population standard deviation, σ = 2.5

Initially, we formulate the null hypothesis and the alternative hypothesis

To conduct this test, we utilize the One-tailed z test.

a) Equation:

z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }

By substituting in all relevant values, we determine:

z_{stat} = \displaystyle\frac{10.2 - 8.9}{\frac{2.5}{\sqrt{6}} } = 1.28

Next, z_{critical} \text{ at 0.05 level of significance } = 1.64

b) The p-value is computed using the z-table.

P-value = 0.1003

The p-value surpasses the alpha of 0.05

c) Because the p-value exceeds the alpha threshold, there is insufficient evidence to dismiss the null hypothesis, thereby supporting the null hypothesis.

Consequently, it is concluded that the authorship of the sonnets belongs to a particular Elizabethan poet.

6 0
1 month ago
P=100a÷t solve for a
tester [12383]

a=1/100pt is the right solution. Initially, you need to multiply both sides of the equation by t, which results in ⇒ pt=100a. After that, inverting the equation gives ⇒100a=pt. Alternatively, dividing both sides of the equation by one hundred (100) also leads us to \frac{100a}{100}=\frac{pt}{100}. Ultimately, the final answer is a=1/100pt is the result. I hope this offers assistance! Thank you for asking your question, and enjoy your day. -Charlie

8 0
3 months ago
Read 2 more answers
How many weeks of flea treatment would Jim’s dog get from one package if each treatment only lasted 3 weeks In a normal year whe
Inessa [12570]

Solution

The number of weeks in a year is 52 weeks

In a typical year, where each flea treatment lasts 4 weeks, Jim's dog would receive

\frac{52}{4} =13 times

Conversely, during especially bad flea years, when treatments only last for 3 weeks, Jim's dog would need

So in a year, Jim's dog would need 17- 13 =4 extra treatments.

4 treatments at 3 weeks each total 4×3 =12 weeks of additional treatment

4 0
1 month ago
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