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3 years ago

Solve for t. You must write your answer in fully simplified form. -7t = 18

Mathematics

1 answer:

Musya8 [376]3 years ago

4 0

Answer: t = -18/7 = 2.57

Step-by-step explanation:

-Solve for t:

$-7t=18$

$\frac{-7t}{-7} = \frac{18}{-7}$

$t=-\frac{18}{7} =2.57$

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What is the missing number ?? /6 =25/30

matrenka [14]

Answer:

Step-by-step explanation:

$\frac{25}{30}$ ( divide numerator and denominator by 5 ) , then

$\frac{25}{30}$ = $\frac{5}{6}$ ( with missing number ? = 5 )

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2 years ago

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3 years ago

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A credit reporting agency claims that the mean credit card debt in a town is greater than $3500. A random sample of the credit c

Lera25 [3.4K]

Answer:

Claim is false

Step-by-step explanation:

Claim : A credit reporting agency claims that the mean credit card debt in a town is greater than $3500.

$H_0:\mu \leq 3500\\H_a:\mu = 3500$

n = 20

Since n <30

So we will use t test

Formula : $t =\frac{x-\mu}{\frac{s}{\sqrt{n}}}$

s = standard deviation = 391

x = 3600

n = 20

$t =\frac{3600-3500}{\frac{391}{\sqrt{20}}}$

$t =1.14$

Degree of freedom = n-1 = 20-1 = 19

α=0.10

So, using t table

$t_({\frac{\alpha}{2},d.f.})$ = 1.72

t critical > t calculated

So we accept the null hypothesis

Hence we reject the claim that the mean credit card debt in a town is greater than $3500.

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3 years ago

What equation best models this data?(use y to represent the population of rabbits and t to represent the year, assuming that 201

liraira [26]

If we see the data closely, a pattern emerges. The pattern is that the ratio of the population of every consecutive year to the present year is 1.6

Let us check it using a couple of examples.

The rabbit population in the year 2010 is 50. The population increases to 80 the next year (2011). Now, $\frac{80}{50}=1.6$

Likewise, the rabbit population in the year 2011 is 80. The population increases to 128 the next year (2012). Again, $\frac{128}{80}$ $=1.6$

We can verify the same ratio with all the data provided.

Thus, we know that the population in any given year is 1.6 times the population of the previous year. This is a classic case of a compounding problem. We know that the formula for compounding is as:

$F=P\times r^n$

Where $F$ is the future value of the rabbit population in any given year

$P$ is the rabbit population in the year "0" (that is the starting year 2010) and that is 50 in this question. (please note that there is just one starting year).