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1 year ago

Select the correct answer.

Mathematics

2 answers:

lakkis [162]1 year ago

8 0

Answer:

6x + 4y ≤ 50

x + y ≥ 10

Step-by-step explanation:

$6 is the cost of stuffed animals $4 is the cost of toy trucks and the her maximum budget is $50 it would be 6x+4y is less then or equal to 50 and there is AT LEAST 10 people so the amounts which are x and y would be equal to or greater than 10.

olga2289 [7]1 year ago

8 0

Answer:

$6x+4y\leq 50$

$x+y\geq 10$

Step-by-step explanation:

$x$ represents the number of stuffed animals.

$y$ represents the number of toy trucks.

Now, according to the problem, she found stuffed animals for $6.00 each and toy trucks for $4.00 each. Also, Laura has a restricted budget of $50. All these can be expressde as

$6x+4y\leq 50$

Also, the problem states that Laura wants to provide one partu favor per person to at least 10 people, that means

$x+y\geq 10$ , because "at least" represents a minimum value, which is expressed with $\geq$ .

Therefore, the inequality system that models this situation is

$6x+4y\leq 50$

$x+y\geq 10$

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A web-based company has a goal of processing 90 percent of its orders on the same day they are received. If 434 out of the next

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Answer:

We conclude that a web-based company are not exceeding their goal of 90%.

Step-by-step explanation:

We are given that a web-based company has a goal of processing 90 percent of its orders on the same day they are received.

434 out of the next 471 orders are processed on the same day.

Let p = proportion of orders processing on the same day they are received.

SO, Null Hypothesis, $H_0$ : p $\leq$ 0.90 {means that they are not exceeding their goal of 90%}

Alternate Hypothesis, $H_0$ : p > 0.90 {means that they are exceeding their goal of 90%}

The test statistics that would be used here One-sample z test for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of orders that are processed on the same day = $\frac{434}{471}$ = 0.92

n = sample of orders = 471

So, the test statistics = $\frac{0.92-0.90}{\sqrt{\frac{0.92(1-0.92)}{471} } }$

= 1.599

The value of z test statistics is 1.599.

Now, at 0.025 significance level the z table gives critical value of 1.96 for right-tailed test.

Since our test statistic is less than the critical value of z as 1.599 < 1.96, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that a web-based company are not exceeding their goal of 90%.

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