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2 years ago

11

Solve this so you get coordinates as ur answer

2 answers:

tamaranim1 [39]2 years ago

4 0

Answer:

$\begin{bmatrix}x+5y=-4\\ 2x+10y=-8\end{bmatrix}$

First, Isolate "x" for x+5y=-4= x=-4-5y

$\begin{bmatrix}2\left(-4-5y\right)+10y=-8\end{bmatrix}$

Simplify

$\begin{bmatrix}-8=-8\end{bmatrix}$

Therefore, the solutions are x=-4-5y

~

Nostrana [21]2 years ago

3 0

Answer: What topic is this again??

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The brand manager for a brand toothpaste must plan a campaign designed to increase brand recognition. He wants to first determin

bulgar [2K]

We need to find out how many adults must the brand manager survey in order to be 90% confident that his estimate is within five percentage points of the true population percentage.

From the given data we know that our confidence level is 90%. From Standard Normal Table we know that the critical level at 90% confidence level is 1.645. In other words, $Z_{Critical}=1.645$ .

We also know that E=5% or E=0.05

Also, since, $\hat{p}$ is not given, we will assume that $\hat{p}$ =0.5. This is because, the formula that we use will have $\hat{p}(1-\hat{p})$ in the expression and that will be maximum only when $\hat{p}$ =0.5. (For any other value of $\hat{p}$ , we will get a value less than 0.25. For example if, $\hat{p}$ is 0.4, then $1-\hat{p}=0.6$ and thus, $\hat{p}(1-\hat{p})=0.24$ .).

We will now use the formula

$n=(\frac{Z_{Critical}}{E})^2\hat{p}(1-\hat{p})$

We will now substitute all the data that we have and we will get

$n=(\frac{1.645}{0.05})^2\times0.5(1-0.5)$

$n=(32.9)^2\times0.25$

$n=270.6025$

which can approximated to n=271.

So, the brand manager needs a sample size of 271

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3 years ago

The director of a state agency believes that the average starting salary for clerical employees in the state is less than $35 c

MrRissso [65]

Answer:

There is not enough evidence to conclude that average starting salary for clerical employees in the state is less than $35,000 per year.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 35,000 dollars per year

Sample mean, $\bar{x}$ = 34,700 dollars per year

Sample size, n = 100

Alpha, α = 0.10

Population standard deviation, σ = 3,500 dollars per year

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 35,000\text{ dollars per year}\\H_A: \mu < 35,000\text{ dollars per year}$

We use one-tailed z test to perform this hypothesis.

Formula:

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

Putting all the values, we have

$z_{stat} = \displaystyle\frac{34700 - 35000}{\frac{3500}{\sqrt{100}} } = -0.8571$

Now, $z_{critical} \text{ at 0.10 level of significance } = -1.28$

Since,

$z_{stat} > z_{critical}$

We fail to reject the null hypothesis and accept the null hypothesis.

Thus, there is not enough evidence to conclude that average starting salary for clerical employees in the state is less than $35,000 per year.

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3 years ago

A lawyer has an annual income of $240,000. The income tax that doctor has to pay is 6%. What is the amount of income tax in doll

ZanzabumX [31]

Answer:

Step-by-step explanation:

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the dr has to be taxes 14,400 dollars

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3 years ago

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Mariulka [41]

Answer:

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3 years ago

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Zolol [24]

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3 years ago

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