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2 years ago

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Claudia makes an initial deposit of $50 into her bank account. She deposits $20 into the account each week. Let x represent the

number of weeks Claudia makes a deposit, and let y represent the amount of money in her account. Which equation represents the amount of money in Claudia's account after x weeks?
y=20x−50

y=−20x+50

y = 50x + 20

y = 20x + 50
WILL MAKE BRAINLIEST ANSWER

1 answer:

Margarita [4]2 years ago

6 0

The answer would be D. because she has 50 dollars in her account and she adds 20 dollars for x amount of weeks so that will get you y=20x+50

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Ray Np is an angle bisector of angle MNQ and measure of Angle PNQ = 2x+1. Find the measure of angle MNQ. If The measure of MNQ=x

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Refer to the diagram below.

Because ray NP bisects ∠MNQ, therefore
∠MNP = ∠PNQ = 2x + 1.
Therefore
∠MNQ = 2*∠PNQ = 2(2x + 1) = 4x + 2.

Because ∠MNQ is given as x² - 10, therefore
x² - 10 = 4x + 2
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∠MNQ = 4*(-2) + 2 = -6°
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Answer: x = 6, and ∠MNQ = 26°

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2 years ago

The horsepower , H(s), required for a racecar to overcome wind resistance is given by the function : H(s) = 0.003s^2+0.07s-0.027

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Average rate of change = [H(100) - H(80)] / (100 - 80)
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2 years ago

Read 2 more answers

In a completely randomized experimental design involving three assembly methods, 30 employees were randomly selected and 10 were

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Answer:

$F = \frac{MSR}{MSE} =\frac{45.89}{6.27}=7.32$

So then the best option is:

a. 7.32

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

If we assume that we have $3$ groups and on each group from $j=1,\dots,10$ we have $10$ individuals on each group we can define the following formulas of variation:

$SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2$

$SS_{between=Treatment}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2$

$SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2$

And we have this property

$SST=SS_{between}+SS_{within}$

Where SST represent the total sum of squares.

The degrees of freedom for the numerator on this case is given by $df_{num}=k-1=3-1=2$ where k =3 represent the number of groups.

The degrees of freedom for the denominator on this case is given by $df_{den}=df_{between}=N-K=30-3=27$ .

And the total degrees of freedom would be $df=N-1=30 -1 =29$

From the info given we know that $MSR=\frac{SSR}{2}=45.89$

And $MSE=\frac{SSE}{27}=6.27$

From definition the F statisitc is defined as:

$F = \frac{MSR}{MSE} =\frac{45.89}{6.27}=7.32$

So then the best option is:

a. 7.32

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2 years ago

A television station plans to ask a random sample of 400 city residents if they can name the news anchor on the evening news at

hodyreva [135]

Answer:

There is an 8.38% probability that the anchor will be fired.

Step-by-step explanation:

For each resident, there is only two possible outcomes. Either they can name the news anchor on the evening news at their station, or they cannot.

This means that the binomial probability distribution will be used in our solution.

However, we are working with samples that are considerably big. So i am going to aaproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X), \sigma = \sqrt{V(X)}$ .

In this problem, we have that:

400 city residents are going to be asked. So $n = 400$ .

Suppose that in fact 12% of city residents could name the anchor if asked. This means that $p = 0.12$ .

So,

$\mu = E(X) = 400*0.12 = 48$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{400*0.12*0.88} = 6.5$ .

They plan to fire the news anchor if fewer than 10% of the residents in the sample can do so.

What is the approximate probability that the anchor will be fired?

10% of the residents is 0.10*400 = 40 residents.

Fewer than 10% is 39 residents. So the probability that the anchor will be fired is the pvalue of Z when $X = 39$ .

So:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{39 - 48}{6.5}$

$Z = -1.38$

$Z = -1.38$ has a pvalue of 0.0838. This means that there is an 8.38% probability that the anchor will be fired.

5 0

2 years ago