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2 years ago

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Sam and Annika ride their bikes to school. Sam rides his bike (k) kilometers . Annika rides her bike 2 less than 4 times as many

kilometers as Sam rides.
Write an expression to represent how many kilometers Annika rides her bike to school?

Please please please help

1 answer:

BaLLatris [955]2 years ago

6 0

4(k-2) when it says 2 less than it means you are subtracting from the constant and 4 times means you are multiplying the whole equation with 4.

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Sophie [7]

<u>Part 1</u>

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We need to make sure the radical is defined, meaning the radicand has to be non-negative. Thus, the domain is $\boxed{[0, \infty)}$

<u>Part 2</u>

<u /> $(g \circ f)(x)=g(4x+1)=\sqrt{4x+1}$

We need to make sure the radical is defined, meaning the radicand has to be non-negative. Thus,

$4x+1 \geq 0\\\\4x \geq -1\\\\x \geq -\frac{1}{4}$

Thus, the domain in interval notation is $\boxed{\left[-\frac{1}{4}, \infty)}$

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1 year ago

In the past, 21% of all homes with a stay-at-home parent, the father is the stay-at-home parent. An independent research firm ha

Afina-wow [57]

Answer:

A sample size of 79 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$\pi = 0.21$

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

What sample size is needed if the research firm's goal is to estimate the current proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a margin of error of 0.09?

A sample size of n is needed.

n is found when M = 0.09. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.09 = 1.96\sqrt{\frac{0.21*0.79}{n}}$

$0.09\sqrt{n} = 1.96\sqrt{0.21*0.79}$

$\sqrt{n} = \frac{1.96\sqrt{0.21*0.79}}{0.09}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.21*0.79}}{0.09})^{2}$

$n = 78.68$

Rounding up to the nearest whole number.

A sample size of 79 is needed.

4 0

3 years ago

How do you do these questions?

Tanzania [10]

Answer:

μ ≈ 2.33

σ ≈ 1.25

Step-by-step explanation:

Each person has equal probability of ⅓.

$\left[\begin{array}{cc}X&P(X)\\1&\frac{1}{3}\\2&\frac{1}{3}\\4&\frac{1}{3}\end{array}\right]$

The mean is the expected value:

μ = E(X) = ∑ X P(X)

μ = (1) (⅓) + (2) (⅓) + (4) (⅓)

μ = ⁷/₃

The standard deviation is:

σ² = ∑ (X−μ)² P(X)

σ² = (1 − ⁷/₃)² (⅓) + (2 − ⁷/₃)² (⅓) + (4 − ⁷/₃)² (⅓)

σ² = ¹⁴/₉

σ ≈ 1.25

3 0

3 years ago

4<br> It takes Naomi 4 minutes to run 5 laps.<br> How many laps per minute is that?<br> Ebon

attashe74 [19]

Answer:

1.25 per min

Step-by-step explanation:

divide 5÷4 and you get 1.25 per min

6 0

2 years ago

The average score of all pro golfers for a particular course has a mean of 70 and a standard deviation of 3.0 (Think of these as

bija089 [108]

Answer:

0.477 is the probability that the average score of the 36 golfers was between 70 and 71.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 70

Standard Deviation, σ = 3

Sample size, n = 36

Let the average score of all pro golfers follow a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(score of the 36 golfers was between 70 and 71)

$\text{Standard error of sampling} = \displaystyle\frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{36}} = \frac{1}{2}$

$P(70 \leq x \leq 71) = P(\displaystyle\frac{70 - 70}{\frac{3}{\sqrt{36}}} \leq z \leq \displaystyle\frac{71-70}{\frac{3}{\sqrt{36}}}) = P(0 \leq z \leq 2)\\\\= P(z \leq 2) - P(z \leq 0)\\= 0.977 - 0.500 = 0.477= 47.7\%$

$P(70 \leq x \leq 71) = 47.7\%$

0.477 is the probability that the average score of the 36 golfers was between 70 and 71.

8 0

3 years ago