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3 years ago

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The fish tank at Matthew's house holds 468 gallons but is leaking at a rate of 4 gallons per day. Ariana's fish tank holds 624 g

allons but is leaking at a rate of 7 gallons per day. After how many days will the amount of water in the two tanks be the same?

1 answer:

user100 [1]3 years ago

6 0

468 plus 624 times 7 divided by 4 and you should get the amount of water in the two tanks

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A trapezoid has bases that measure 10cm and 6cm. The height of the figure is 15cm. What is the area of the trapezoid?

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Hello!

To find the area of a trapezoid you use the equation

$A = \frac{a+b}{2} h$

A is area
h is height
a and b are the bases

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$A = \frac{10+6}{2} *15$

Add

$A = \frac{16}{2} *15$

Divide

$A = 8 * 15$

Multiply

A = 120

The answer is 120cm squared

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Fill in the blanks using a variable or variables to rewrite the given statement. Is there an integer that has a remainder of 2 w

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Answer:

Step-by-step explanation:

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A notebook has a perimeter of 36 inches and an area of 80 square inches. What are the dimensions of the notebook?

maw [93]

Let L be length and W be width
The perimeter of a rectangle is 2W+2L=36
Thus 2W=36-2L, W=18-L
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Thus (18-L)(L)=80
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3 years ago

The proportion of defective computers built by Byte Computer Corporation is 0.15. In an attempt to lower the defective rate, the

FinnZ [79.3K]

Answer:

a)<em> Null hypothesis : H₀</em>: the proportion of defective item of computer has been lowered. That is P < 0.15

<u><em>Alternative hypothesis: H₁:</em></u> The proportion of defective item of computer

has been higher. That is P> 0.15 (Right tailed test)

b) Test statistic $Z = \frac{p-P}{\sqrt{\frac{PQ}{n} } }$

c) Calculate the value of the test statistic = 0.991

d) The critical value at 0.01 level of significance = Z₀.₀₁ = 2.57

e) Null hypothesis accepted at 0.01 level of significance

f) we accepted null hypothesis.

Hence t<em>he proportion of defective item of computer has been lowered. </em>

Step-by-step explanation:

<u>Step(i)</u>:-

<em>Given the sample size 'n' = 42</em>

Given random sample of 42 computers were tested revealing a total of 4 defective computers.

The defective computers 'x' = 4

<em>The sample proportion of defective computers </em>

$p = \frac{x}{n} = \frac{4}{42} = 0.095$

<em>Given The Population proportion 'P' = 0.15</em>

<em>The level of significance ∝=0.01</em>

<u>Step(ii)</u>:-

a)<em> Null hypothesis : H₀</em>: the proportion of defective item of computer has been lowered. That is P < 0.15

<u><em>Alternative hypothesis: H₁:</em></u> The proportion of defective item of computer

has been higher. That is P> 0.15 (Right tailed test)

b)

Test statistic $Z = \frac{p-P}{\sqrt{\frac{PQ}{n} } }$

c)

$Z = \frac{0.095-0.15}{\sqrt{\frac{0.15(0.85)}{42} } }$

$z = \frac{-0.055}{\sqrt{0.00303} } = - 0.9991$

Calculate the value of the test statistic Z = - 0.9991

|Z| = |- 0.9991| = 0.991

<u>Step(iii)</u>:-

d)

The critical value at 0.01 level of significance = Z₀.₀₁ = 2.57

e) Calculate the value of the test statistic Z = 0.991 < 2.57 at 0.01 level of significance.

<u><em>Conclusion</em></u>:-

Hence the null hypothesis is accepted at 0.01 level of significance.

f)

<em> The proportion of defective item of computer has been lowered.</em>

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3 years ago