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matrenka [14]
4 years ago
7

Please help with these two math problems 1. 7^x ÷ 7^y = 2. z^10x^y ÷ z^5 =

Mathematics
1 answer:
IceJOKER [234]4 years ago
8 0

Answer:

  1.  7^(x-y)

  2. z^5x^y

Step-by-step explanation:

The applicable rule of exponents is ...

  (a^b)/(a^c) = a^(b-c)

__

1.

  \dfrac{7^x}{7^y}=\boxed{7^{x-y}}

__

2.

  \dfrac{z^{10}x^y}{z^5}=x^yz^{10-5}=\boxed{z^5x^y}

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bazaltina [42]

Answer:

2:25

Step-by-step explanation:

6 0
3 years ago
I was wondering what is 9/2 divided by 1/2
liubo4ka [24]

Answer:

\frac{18}{2} or 9

Step-by-step explanation:

1. Set up the problem: \frac{9}{2} ÷ \frac{1}{2}

2. Replace the division sign with a multiplication sign: \frac{9}{2} × \frac{1}{2}

3. Find the reciprocal by swapping the numerator and denominator on the second fraction:  \frac{9}{2} × \frac{2}{1}

4. Multiply fractions:  \frac{9}{2} × \frac{2}{1} = \frac{18}{2} or 9

6 0
3 years ago
A manufacturer of metal fasteners expects to ship an average of 1197.00 boxes of fasteners per day. A random sample of 24 days p
lana66690 [7]

Answer:

The p-value obtained is less than the significance level at which the test was performed, hence, we reject the null hypothesis, accept the alternative hypothesis & conclude that there is evidence showing that the average number of boxes shipped per day is different from 1197.00

Step-by-step explanation:

For hypothesis testing, we first clearly state our null and alternative hypothesis.

The null hypothesis is that there is no evidence showing that the average number of boxes shipped per day is different from 1197.00. That is, there is no significant difference between the number of boxes shipped per day and 1197.00

And the alternative hypothesis is that there is evidence showing that the average number of boxes shipped per day is different from 1197.00

Mathematically, the null hypothesis is

H₀: μ₀ = 1197.00

The alternative hypothesis is

Hₐ: μ₀ ≠ 1197.00

To do this test, we will use the t-distribution because no information on the population standard deviation is known

So, we compute the t-test statistic

t = (x - μ₀)/σₓ

x = sample mean = 1193.69 boxes per day

μ₀ = standard that we're comparing the sample mean with = 1197.00

σₓ = standard error of the sample mean

= (σ/√n)

where n = Sample size = 24 days

σ = standard deviation of the sample = 3.3166 boxes per day

σₓ = (3.3166/√24) = 0.677

t = (1193.69 - 1197) ÷ 0.677

t = -4.89

checking the tables for the p-value of this t-statistic

Degree of freedom = df = n - 1 = 24 - 1 = 23

Significance level = 0.05

p-value (for t = -4.89, at 0.05 significance level, with a two tailed condition) = 0.000061

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 5% = 0.05

p-value = 0.000061

0.000061 < 0.05

Hence,

p-value < significance level

This means that we reject the null hypothesis, accept the alternative hypothesis & conclude that there is evidence showing that the average number of boxes shipped per day is different from 1197.00

Hope this Helps!!!

6 0
3 years ago
After constructing a relative frequency distribution summarizing IQ scores of college students, what should be the sum of the re
saul85 [17]

Answer:

  • <u>Option</u><u> </u><u><em>A</em></u><em>. If percentages are used, the sum should be a 100%. If proportions are used, the sum should be 1.</em>

Explanation:

The <em>relative frequency</em> can be measured as a ratio (proportion) or as a percentage.

As a ratio, the relative frequency is the number of times that the desired outcome is observed (either theoretically or experimentally) divided by the total number of  possible outcomes (theoretically) or  observed (experimentally).

As a<em> percentage</em>, the relative frequency is the ratio multiplied by 100.

This shows it mathematically:

  • relative frequency of event 1 = frequency of event 1 / number of events

  • relative frequency of event 2 = frequency of event 2/number of events

  • relative frequency of event n = freqquency of event n/number of events

  • Total relative frequency = sum of of relative frequencies

                                       = sum of all (n) frequencies / number of events =

                                       = number of events / number of events = 1

As a percentage, total relative frequency = 1 × 100 = 100

7 0
4 years ago
Y is directly proportional to x.
goldenfox [79]
A) it would be a ratio of 5:1, y=5x
B) if x is 12, it has doubled, so you do the same to y, so it would be 60
3 0
4 years ago
Read 2 more answers
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