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2 years ago

Which expressions are equivalent to 7^{-2}\cdot 7^67

Mathematics

2 answers:

AleksandrR [38]2 years ago

3 0

Answer:

${7}^{ - 2} \cdot \: {7}^{6} = {7}^{ 4}$

Step-by-step explanation:

The given expresion is;

${7}^{ - 2} \cdot {7}^{6}$

We apply the product rule of indices;

${a}^{m} \cdot \: {a}^{n} = {a}^{m + n}$

We apply this property to our expression to get:

${7}^{ - 2} \cdot \: {7}^{6} = {7}^{ - 2 + 6}$

Simplify the exponent on the right;

${7}^{ - 2} \cdot \: {7}^{6} = {7}^{ 4}$

We can't see the options, but I hope this helps!

MrRa [10]2 years ago

3 0

Answer:

7^2/7^-2 and (7^2)^2

Step-by-step explanation:

please mark me as Brainliest

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A cell tower is located 3 miles east and 4 miles north of the center of a small town. The cell tower has a coverage radius of 3

butalik [34]

Answer:

The answer is below

Step-by-step explanation:

A cell tower is located 3 miles east and 4 miles north of the center of a small town. The cell tower has a coverage radius of 3 miles.

a) Start by drawing a diagram of this situation. Your diagram might include a coordinate plane, the cell tower, and a circle representing the cell tower's coverage boundary.

a) A house is located 5 miles north of the center of the town and is to the east of the cell tower. If the house lies on the boundary of the cell tower's coverage, how far east of the center of the town is the house?

Answer:

Let the center of the town represent the origin (0,0). Also 1 unit = 1 mile. Since the cell tower is located 3 miles east and 4 miles north of the center of a small town, it is represented by A(3, 4)

The cell tower has a coverage of radius 3 miles. This can be represented by a circle with equation:

(x - a)² + (y - b)² = r². where (a,b) is the center of the circle and r is the radius. Hence:

(x - 3)² + (y - 4)² = 3²

(x - 3)² + (y - 4)² = 9

The diagram is drawn using geogebra.

b) The house is 5 miles north. It can be represented by y = 5 line.

To find the distance east of the house we have to substitute y = 5 and solve for x, hence:

(x - 3)² + (y - 4)² = 9

(x - 3)² + (5 - 4)² = 9

(x - 3)² + 1 = 9

(x - 3)² = 8

(x - 3) = √8

x - 3 = ±2.83

x = 3 ± 2.83

x = 5.83 or 1.83

Since it is to the east to the cell tower, hence x = 5.83.

Therefore the house is located 5.83 miles to the east of the cell tower

8 0

2 years ago

In Juneau, Alaska, the 30-year annual snowfall average is 86.7 inches with a standard deviation of 40.4 inches. The last four ye

marusya05 [52]

The expected value of the amount of average snowfall for over 30 years is 86.7 inches with a standard deviation of 40.4 inches. To verify if this particular trend continues, we must check the significance value of the amount snowfall for the past four years. Given that the snowfall for past years are as follows: 115.7 inches, 62.9 inches, 168.5 inches, and 135.7 inches. Thus the mean of the sample would be: (115.7 + 62.9 + 168.5 + 135.7)/4 = 120.7 inches. To compute for the z-score, we have z-score = (x – μ) / (σ / √n) where x is the computed/measured value, μ is the expected mean, σ is the standard deviation, and n is the number of samples. Using the information we have, z-score (z) = (120.7 - 86.7) / (40.4/ √4) = 1.68 In order to reject the null hyptohesis our probability value must be less than the significance level of 5%. For our case, since z = 1.68, P-value = 0.093 > 0.05. Therefore, the answer is B.

5 0

2 years ago

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Last year, 46% of business owners gave a holiday gift to their employees. A survey of business owners indicated that 35% plan to

DiKsa [7]

Answer:

a) Number = 60 *0.35=21

b) Since is a left tailed test the p value would be:

$p_v =P(Z$

c) If we compare the p value obtained and using 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% the proportion of business owners providing holiday gifts had decreased from the 2008 level.

We can use as smallest significance level 0.044 and we got the same conclusion.

Step-by-step explanation:

Data given and notation

n=60 represent the random sample taken

X represent the business owners plan to provide a holiday gift to their employees

$\hat p=0.35$ estimated proportion of business owners plan to provide a holiday gift to their employees

$p_o=0.46$ 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)

Part a

On this case w ejust need to multiply the value of th sample size by the proportion given like this:

Number = 60 *0.35=21

2) Part b

We need to conduct a hypothesis in order to test the claim that the proportion of business owners providing holiday gifts had decreased from the 2008 level:

Null hypothesis: $p\geq 0.46$

Alternative hypothesis: $p < 0.46$

When we conduct a proportion test we need to use the z statisitc, 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 $p_o$ .

Calculate the statistic

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

$z=\frac{0.35 -0.46}{\sqrt{\frac{0.46(1-0.46)}{60}}}=-1.710$

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 left tailed test the p value would be:

$p_v =P(Z$

Part c

If we compare the p value obtained and using 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% the proportion of business owners providing holiday gifts had decreased from the 2008 level.

We can use as smallest significance level 0.044 and we got the same conclusion.

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1 year ago

You're at a clothing store that dyes your clothes while you wait. You get to pick from 4 pieces of clothing (shirt, pants, socks

nataly862011 [7]

Answer:

¼ chance

Step-by-step explanation:

If there is 4 items and 1 blue die and its a ¼ chance if there was more blue then there would be a higher chance for having blue socks

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1 year ago

Kelly wants to know the actual value of a house that she wants to buy. What should she do?

Ghella [55]

Kelly should ask the realtor.

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1 year ago

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