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zimovet [89]
3 years ago
6

As an estimation we are told ?3 is €4. Convert €83.60 to pounds. Give your answer rounded to 2 DP.

Mathematics
1 answer:
bija089 [108]3 years ago
7 0

Answer:

€83.60 = £62.70

Step-by-step explanation:

£3/€4 = £x/€ 83.60

¾ = x/83.60

Multiply each side by the lowest common denominator (4 × 83.60)

3 × 83.60 = 4x

Divide each side by 4

x = (3 × 83.60)/4 = £62.70

∴ €83.60 = £62.70

(The conversion factor is out of date. The current conversion factor is closer to £6 = €7.)

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Side MN = 6, side OM = 9, side PQ = 6, and side RP = 9. What side corresponds to side NO and can be used to show that ΔMNO ≅ ΔPQ
kirill115 [55]
If triangle MNO is congruent to triangle PQR then side OM is congruent to side RP and equals 9
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3 years ago
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Write an equation for the line perpendicular to the line y= -2/3+7 which passes through the point (-5,6)
zysi [14]
Y = -2/3x + 7.....the slope here is -2/3. A perpendicular line will have a negative reciprocal slope. All tht means is take the original slope, flip it, and change the sign. So we take -2/3....flip it making it -3/2.....change the sign making it 3/2. So ur perpendicular line will have a slope of 3/2.

y = mx + b
slope(m) = 3/2
(-5,6)....x = -5 and y = 6
now sub and find b, the y int
6 = 3/2(-5) + b
6 = -15/2 + b
6 + 15/2 = b
12/2 + 15/2 = b
27/2 = b
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2 years ago
Which number is not equal to one of the following expressions?
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C: 703.5 is the answer
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3 years ago
Use this information to answer the questions. University personnel are concerned about the sleeping habits of students and the n
Oksanka [162]

Answer:

z=\frac{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097  

p_v =P(Z>2.097)=0.018  

If we compare the p value obtained and the significance level given \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  students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

Step-by-step explanation:

1) Data given and notation

n=377 represent the random sample taken

X=209 represent the students reported experiencing excessive daytime sleepiness (EDS)

\hat p=\frac{209}{377}=0.554 estimated proportion of students reported experiencing excessive daytime sleepiness (EDS)

p_o=0.5 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.5:  

Null hypothesis:p\leq 0.5  

Alternative hypothesis:p > 0.5  

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.

3) Calculate the statistic  

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

z=\frac{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097  

4) 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 significance level provided \alpha=0.05. 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>2.097)=0.018  

If we compare the p value obtained and the significance level given \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  students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

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3 years ago
Find x. and rounddddd
Lerok [7]

Answer:

X=117.9

Step-by-step explanation:

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