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ValentinkaMS [17]
3 years ago
7

Sandra scores 12 out of 30 in her maths test. Write this as a fraction in its simplest form.​

Mathematics
1 answer:
Andrew [12]3 years ago
8 0

Step-by-step explanation:

12/30

find highest common factor; 6

divide both sides by 6

(12/6) (30/6)

2/5

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I don't get it- can someone explain
labwork [276]

Answer:

11 4-passenger cars

Step-by-step explanation:

let 'x' represent the number of 4-passenger cars,then 'x-3' represent the number of 6-passenger cars.

92=4x+6(x-3)

92=4x+6x-18

92=10x-18

92+18=10x

110=10x

x=110/10

x=11

Therefore there are 11 4-passenger cars

8 0
3 years ago
What is the range of the function graphed below?
I am Lyosha [343]

Answer:

The range of the graph is:

-∞ <  y < 2

Hence, option C is correct.

Step-by-step explanation:

We also know that range is the set of values of the dependent variable for which a function is defined.

In other words,

Range refers to all the possible sets of output values on the y-axis.

From the given graph, we can easily observe the vertical extent of the graph is all range values  between -∞ and 2.

In other words,

The range of the function lies in the interval: ( -∞, 2)

Therefore, the range of the graph is:

-∞ <  y < 2

Hence, option C is correct.

7 0
3 years ago
Which relationship in the triangle must be true?
forsale [732]

Answer:

sinB=cos(90-B)

Step-by-step explanation:

this is the relationship between sin and cos. other choices are wrong

3 0
3 years ago
Unlike most packaged food products, alcohol beverage container labels are not required to show calorie or nutrient content. An a
zhenek [66]

Answer:

We conclude that the true average estimated calorie content in the population sampled exceeds the actual content.

Step-by-step explanation:

We are given that an article reported on a pilot study in which each of 58 individuals in a sample was asked to estimate the calorie content of a 12 oz can of beer known to contain 153 calories.

The resulting sample mean estimated calorie level was 193 and the sample standard deviation was 88.

Let \mu = <u><em>true average estimated calorie content in the population sampled.</em></u>

So, Null Hypothesis, H_0 : \mu \leq 153 calories     {means that the true average estimated calorie content in the population sampled does not exceeds the actual content}

Alternate Hypothesis, H_A : \mu > 153 calories     {means that the true average estimated calorie content in the population sampled exceeds the actual content}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

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

where, \bar X = sample mean estimated calorie level = 193 calories

            s = sample standard deviation = 88

            n = sample of individuals = 58

So, <u><em>the test statistics</em></u>  =  \frac{193-153}{\frac{88}{\sqrt{58} } }  ~ t_5_7

                                       =  3.462

The value of t test statistics is 3.462.

Since, in the question we are not given the level of significance so we assume it to be 5%. <u>Now, at 0.05 significance level the t table gives critical value of 1.6725 at 57 degree of freedom for right-tailed test.</u>

Since our test statistic is more than the critical value of t as 3.462 > 1.6725, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the true average estimated calorie content in the population sampled exceeds the actual content.

8 0
3 years ago
I need help, please, I'm not that good at math
exis [7]

Answer:

Please help me

Step-by-step explanation:

8 0
3 years ago
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