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NARA [144]
2 years ago
7

The goal of a toy drive is to donate more than 1000 toys. The toy drive already has collected 300 toys. How many more toys does

the toy drive need to meet its goal? Write and solve an inequality shows the points you must earn to go the next level?
Mathematics
1 answer:
Brilliant_brown [7]2 years ago
7 0

Answer:

Let t be the number of toys.

Since, the toy drive already has collected 300 toys

⇒ t+ 300

Also, it is given that the goal of a toy drive is to donate more than 1000 toys.

We get an inequality:

⇒t + 300 > 1000

Solve an inequality:  

t + 300 > 1000

Subtract 300 from both sides we get

t + 300 -300 > 1000-300

Simplify:

t > 700

Therefore, more than 700 toys does the toy drive need to meet its goal.

An inequality is; t + 300 > 1000 ;  t > 700




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The probability that a randomly selected elementary or secondary school teacher from a city is a female is , holds a second job
julsineya [31]

Answer:

The probability that an elementary or secondary school teacher selected at random from this city is a female or holds a second job is 0.90.

Step-by-step explanation:

Denote the events as follows:

<em>X</em> = an elementary or secondary school teacher from a city is a female

<em>Y</em> = an elementary or secondary school teacher holds a second job

The information provided is:

P (X) = 0.66

P (Y) = 0.46

P (X ∩ Y) = 0.22

The addition rule of probability is:

P(A\cup B)=P(A)+P(B)-P(A\cap B)

Use this formula to compute the probability that an elementary or secondary school teacher selected at random from this city is a female or holds a second job as follows:

P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\=0.46+0.66-0.22\\=0.90

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3 years ago
Faced with rising fax costs, a firm issued a guideline that transmissions of 10 pages or more should be sent by 2-day mail inste
Vladimir [108]

Answer:

The value of t test statistics is 5.9028.

We conclude that the true mean is greater than 10 at the .01 level of significance.

Step-by-step explanation:

We are given that a firm issued a guideline that transmissions of 10 pages or more should be sent by 2-day mail instead.

The firm examined 35 randomly chosen fax transmissions during the next year, yielding a sample mean of 14.44 with a standard deviation of 4.45 pages.

<em />

<em>Let </em>\mu<em> = true mean transmission of pages.</em>

So, Null Hypothesis, H_0 : \mu \leq 10 pages     {means that the true mean is smaller than or equal to 10}

Alternate Hypothesis, H_A : \mu > 10 pages     {means that the true mean is greater than 10}

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                        T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

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(a) The value of t test statistics is 5.9028.

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Therefore, we conclude that the true mean is greater than 10.

(b) Now, P-value of the test statistics is given by the following formula;

               P-value = P( t_3_4 > 5.9028) = Less than 0.05%    {using t table)

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