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k0ka
8 days ago
14

Antonia is keeping track of the number of pages she reads. She started the school year having already read 95 pages her book, an

d she then reads 20 pages each day. The function for her total pages read is f(x) = 20x + 95. What do f(x) and x represent in Antonia's situation?
Mathematics
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Which of the following functions is graphed below?
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Answer:

Option (B)

Step-by-step explanation:

The given question is incomplete; please refer to the attached document for the full details.

In the attached graph,

The parent function is defined as an absolute value function,

f(x) = |x|

When this graph is moved four units to the left, the translation rule becomes,

f(x) → f(x + 4)

Consequently, the new function following this translation is,

g(x) = f(x + 4) = |x + 4|

Now, with the graph shifted down by two units, the translated function becomes,

h(x) = g(x) - 2

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When reformulated as an equation, the graph can be represented by

⇒ y = |x + 4| - 2

Thus, Option (B) is the correct answer.

8 0
3 months ago
Translate into an algebraic expression and simplify if possible. C It would take Maya x minutes to rake the leaves and Carla y m
lawyer [12517]

Answer:

together they rake \frac{y+x}{xy} leaves in one minute.

Step-by-step explanation:

If Maya takes x minutes to rake, then in one minute, she rakes \frac{1}{x} leaves.

For Carla, if her raking takes y minutes, in one minute she rakes \frac{1}{y} leaves.

Thus, to find the amount they work together in one minute, we add both of their contributions for that minute, resulting in: \frac{1}{x}+ \frac{1}{y}

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2 months ago
An experiment was conducted to record the jumping distances of paper frogs made from construction paper. Based on the sample, th
tester [12383]

Answer:

9.9676 - 2.326*0.5904 =8.594

9.9676 + 2.326*0.5904 =11.341

Step-by-step explanation:

Notation

\bar X is the sample mean

\mu indicates the population mean (the variable of interest)

s signifies the sample standard deviation

n denotes the sample size

Solution to the problem

The mean's confidence interval is derived from the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In this instance, the 9% confidence interval corresponds to:

8.8104 \leq \mu \leq 11.1248

We can determine the mean using the following:

\bar X = \frac{8.8104 +11.1248}{2}= 9.9676

Additionally, the margin of error can be calculated as:

ME= \frac{11.1248- 8.8104}{2}= 1.1572

The margin of error for this situation is expressed as:

ME = t_{\alpha/2}\frac{s}{\sqrt{n}} = t_{\alpha/2} SE

Next, we find the standard error:

SE = \frac{ME}{t_{\alpha/2}}

The critical value for a 95% confidence interval using a normal standard distribution is roughly 1.96, and substituting gives us:

SE = \frac{1.1572}{1.96}= 0.5904

For the 98% confidence interval, the significance corresponds to \alpha=1-0.98= 0.02 and \alpha/2 = 0.01 with a critical value of 2.326, yielding a confidence interval of:

9.9676 - 2.326*0.5904 =8.594

9.9676 + 2.326*0.5904 =11.341

8 0
2 months ago
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