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1 year ago

15

You want to rent a television. Company A charges 50$ per week plus a one time fee of 15$. Company B charges 30$ per week plus a

one time fee of 55$. After how many weeks are the total costs the same at both companies

1 answer:

SCORPION-xisa [38]1 year ago

4 0

Answer: 2 weeks

Step-by-step explanation:

You might be interested in

It’s number 13 and I have to use the area formulas to find each missing measure

ivanzaharov [21]

13)
A = bh
h = A/b
h = 211.4 / 24.3
h = 8.7

answer
height = 8.7 m

6 0

2 years ago

Which measure is the largest? 1.2 km, 120,600 cm, 1,220,000 mm, 120 meters

Margarita [4]

The first thing you should know to answer this question is the following conversions:
1Km = 1000m
1m = 1000mm
1Km = 1000000mm
1m = 100cm
By making the conversions to the same unit, we can know what measure is greater:
1.2Km = 1200m
120600cm = 1206.00m
1220000mm = 1220m
120m = 120m
Answer
The biggest measure is
1220000mm = 1220m

6 0

2 years ago

Suppose that a system of linear equations has 3×5 augmented matrix whose 5th column is a pivot column. Is the system consistent?

Anon25 [30]

No, the system is inconsistent.

Step-by-step explanation:

If the last column is a pivot column, then that row gives an equation that looks something like 0x+0y+0z=1 , meaning , 0=1. Clearly, this is false.

So, the system of linear equations is inconsistent.

7 0

2 years ago

To plan the budget for next year a college must update its estimate of the proportion of next year's freshmen class that will ne

Naily [24]

Answer:

Null hypothesis: $p\leq 0.35$

Alternative hypothesis: $p > 0.35$

$z=\frac{0.447 -0.35}{\sqrt{\frac{0.35(1-0.35)}{150}}}=2.491$

$p_v =P(z>2.491)=0.0064$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of applicants who request financial aid is significantly higher than 0.35

Step-by-step explanation:

Data given and notation

n=150 represent the random sample taken

X=67 represent the applicants who request financial aid

$\hat p=\frac{67}{150}=0.447$ estimated proportion of applicants who request financial aid

$p_o=0.35$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of applicants who request financial aid is higher than 0.35.:

Null hypothesis: $p\leq 0.35$

Alternative hypothesis: $p > 0.35$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.447 -0.35}{\sqrt{\frac{0.35(1-0.35)}{150}}}=2.491$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>2.491)=0.0064$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of applicants who request financial aid is significantly higher than 0.35

5 0

2 years ago

The track team is trying to reduce their time for a relay race. First they reduce their time by 2.1 minutes. Then they are able

konstantin123 [22]

Answer: 15.7 minutes

Step-by-step explanation:

Let x be the time in the beginning (in minutes).

Given: The track team is trying to reduce their time for a relay race.

First they reduce their time by $t_1=$ 2.1 minutes.

Then they are able to reduce that time by $t_2=$ 10

If their final time is 3.96 minutes, then

$x-t_1-t_2=3.6\\\Rightarrow\ x=3.6+t_1+t_2\\\Rightarrow\ x=3.6+2.1+10\\\Rightarrow\ x= 15.7$

Hence, their beginning time was 15.7 minutes.

7 0

2 years ago