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2 years ago

Four friends were born in consecutive years. The sum of their ages is 62. What is the age of the oldest friend?

1 answer:

3 0

Since our sum is 62, and we have consecutive numbers being added, our range is most likely going to be from 10-20.
To further solve this problem easily, try finding 4 consecutive numbers with the ones place adding up to 12, 22, 32, 42, or any number that sums up to the 2 in the ones place.
Let's try 14, 15, 16, 17, since 4, 5, 6, and 7 sum up to a number with a 2 in the ones place.
14 + 15 = 29,
16 + 17 = 33.
33 + 29 is 62.
We have found the age of the friends, but we're looking for the oldest.
Our largest number is 17.
Your answer is 17 years old.
I hope this helps!

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An experiment was performed to compare the wear of two different laminated materials. Twelve pieces of material 1 were tested by

GenaCL600 [577]

Answer:

At 0.05 level of significance, the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Step-by-step explanation:

We hypothesize that mean difference between abrasive wear of material 1 and material 2 is greater than 2.

So we write the null hypothesis $H_0 : \mu_1 - \mu_2 >2$ ,

and the alternative hypothesis $H_1: \mu_1 - \mu_2 \leq 2$ .

We will find the T-score as well as the p-value. If the p-value is less than the level of significance, we will reject the null hypothesis, i.e. we will conclude that the abrasive wear of material 1 is less than that of material 2. Otherwise, we will accept the null hypothesis.

Since the variance is unknown and assumed to be equal, we will use the pooled variance

$s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2} = 20.05$ ,

where $n_1 = 12, n_2 =10, s_1 =4, s_2 = 5$ .

The mean of material 1 and material 2 are $\mu_1 =85, \mu_2=81$ respectively and mean difference $d$ is equal to 4. The hypothesize difference $d_0$ is equal to 2.

To find the T-score, we use the following formula

$T = \frac{d - d_0}{\sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_2} }}$

Substituting all the values into the T-score formula gives us $T = 1.04$ , and the respective p-value is equal to 0.31. This means we have enough statistical evidence not to reject the null hypothesis, and at 5% significance level, the abrasive wear of material 1 exceeds that of material 2 by more than 2 units.

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2 years ago

At the end of the first school day, Mr. Davies, a first grade teacher, inspected his classroom crayons. 2 of the 200 crayons w

MrRissso [65]

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2 years ago

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After simplifying, which expressions are equivalent? Select three options.

Wewaii [24]

Answer: Second option, third option and fifth option.

Step-by-step explanation:

In order to solve this exercise it is important to remember the multiplication of signs:

$(+)(+)=+\\(-)(-)=+\\(-)(+)=-\\(+)(-)=-$

Knowing that, you can distribute the sign and add the like terms. Repeat this procedure in each expression given in the exercise.

Therefore, you get:

$a.\ (3.4a - 1.7b) + (2.5a - 3.9b)=3.4a - 1.7b + 2.5a - 3.9b=5.9a-5.6b\\\\b.\ (2.5a + 1.6b) + (3.4a + 4b)=2.5a + 1.6b + 3.4a + 4b=5.9a+5.6b\\\\c.\ (-3.9b + a) + (-1.7b + 4.9a)=-3.9b + a -1.7b + 4.9a=5.9a-5.6b\\\\d.\ -0.4b +(6b - 5.9a)=-0.4b +6b - 5.9a=6b-6.3a\\\\e.\ 5.9a - 5.6b$

You can identify that the expressions $(2.5a + 1.6b) + (3.4a + 4b)$ , $(-3.9b + a) + (-1.7b + 4.9a)$ and $5.9a - 5.6b$ are equivalent after simplifying.

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2 years ago

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Which of the following slopes show that the set of points C(1, 1), D(3, -4), E(5, 8) are not collinear?

MArishka [77]

C is the answer you want

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2 years ago

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Recently, 18.3% of visitors to the travelocity web site hailed from the google search engine. a random sample of 130 visitors on

kaheart [24]

In your problem:
p = 18.3% = 0.183
n = 130

The standard error can be calculated by the formula:
SE = √[p · (1 - p) / n]
= √[0.183 · (1 - 0.183) / 130]
= 0.0339

The standard error of the proportion is 0.034.

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2 years ago