All categories

3 years ago

What is the product?

Mathematics

1 answer:

Dennis_Churaev [7]3 years ago

8 0

Answer:

84x¹²

Step-by-step explanation:

it is this because when you multiply -7³×-3⁸= 21¹¹ which means when you times 4x you also have to add a 1 to the exponent which makes it 84x¹²

You might be interested in

The amount of money spent on textbooks per year for students is approximately normal.

Ostrovityanka [42]

Answer:a

$336.04 < \mu < 443.96$

The margin of error will increase

The margin of error will decreases

The 99% confidence interval is $0.4107 < p < 0.4293$

Step-by-step explanation:

From the question we are told that

The sample size $n = 19$

The sample mean is $\= x = \$\ 390$

The standard deviation is $\sigma = \$ \ 120$

Given that the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

$Z_{\frac{\alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 1.96 * \frac{120}{\sqrt{19} }$

=> $E = 53.96$

The 95% confidence interval is

$\= x - E < \mu < \= x + E$

=> $390 - 53.96 < \mu < 390 - 53.96$

=> $336.04 < \mu < 443.96$

When the confidence level increases the $Z_{\frac{\alpha }{2} }$ also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

$n \ \ \alpha \ \ \frac{1}{E^2 }$

So if the sample size increases the margin of error decrease

The sample proportion is mathematically represented as

$\r p = \frac{210}{500}$

$\r p = 0.42$

Given that the confidence level is 0.99 the level of significance is $\alpha = 0.01$

The critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} }* \sqrt{ \frac{\r p (1- \r p )}{n} }$

=> $E = 0.42 * \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }$

=> $E = 0.0093$

The 99% confidence interval is

$\r p - E < p < \r p + E$

$0.42 - 0.0093 < p < 0.42 + 0.0093$

$0.4107 < p < 0.4293$

4 0

4 years ago

HELP ME PLEASE, I NEED THIS BY TONIGHT

saul85 [17]

The first option and the third option are correct.

(Although, it's probably way too late now...)

7 0

4 years ago

2 divided by one over 3 the awnser is a fraction

slamgirl [31]

Answer:

That would equal $\frac{2}{3}$ or approximately 0.67. If this helps, please rank Brainliest. Thanks!

4 0

3 years ago

Read 2 more answers

Perry pays $340 per month to lease her car. How much will she pay to lease her car for two years?

irinina [24]

$340 × 24 (12 months in one year)

= $8160

5 0

3 years ago

Read 2 more answers

In a mid-size company, the distribution of the number of phone calls answered each day by each of the 12 receptionists is bell-s

Radda [10]

Answer:

68%

Step-by-step explanation:

The Standard Deviation Rule = Empirical rule formula states that:

68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.

95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.

99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.

From the question,

Step 1

We have to find the number of Standard deviation from the mean. This is represented as x in the formula

μ = Mean = 61

σ = Standard Deviation = 8

For x = 53

μ - xσ

53 = 61 - 8x

8x = 61 - 53

8x = 8

x = 8/8

x = 1

For x = 69

μ + xσ

69 = 61 + 8x

8x = 69 - 61

8x = 8

x = 8/8

x = 1

This falls within 1 standard deviation of the mean where: 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.

Therefore, according to the Standard Deviation Rule, the approximate percentage of daily phone calls numbering between 53 and 69 is 68%

3 0

3 years ago