All categories

3 years ago

Describe and correct the error below.

Mathematics

1 answer:

RUDIKE [14]3 years ago

3 0

Answer:

Software comprises the entire set of programs, procedures, and routines associated with the operation of a computer system. The term was coined to differentiate these instructions from hardware—i.e., the physical components of a computer system.

You might be interested in

The container that holds the water for the football team is 1/2 full. after pouring in 9 gallons oo water, it is 4/5 full. how m

OLEGan [10]

.5x + 9 = .8x
9 = .3x
x = 30 gallons

8 0

3 years ago

Write a quadratic equation whose roots are -4 and one and who's leading coefficient is five

Alex

Step-by-step explanation:

- 4 and 1 are roots.

Therefore, (x + 4) & (x - 1) are factors and 5 is leading coefficient.

Therefore, required quadratic equation is:

$5 (x + 4) (x - 1) \\ = 5( {x}^{2} - x + 4x - 4) \\ = 5( {x}^{2} + 3x - 4) \\ = 5 {x}^{2} + 15x - 20$

7 0

3 years ago

Suppose you do not know the population mean fee charged to H&R Block customers last year. Instead, suppose you take a sample

puteri [66]

Answer:

i $\to$ a

$n = 96040000$

i $\to$ b

$n_1 =24010000$

i $\to$ c

$n_2 =41602500$

ii $\to$ a

$E = 58.16$

ii $\to$ b

$291.84 < \mu < 408.16$ \

ii $\to$ c

There is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval

Step-by-step explanation:

From the question we are told that

The sample size is $n = 8$

The sample mean is $\= x = \$ 350$

The sample standard deviation is $\$ 100$

Considering question i

i $\to$ a

At $E = 0.02$

given that the confidence level is 95% = 0.95

the level of significance would be $\alpha =1-0.95 = 0.05$

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

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

So the sample size is mathematically evaluated as

$n = [ \frac{Z_{\frac{\alpha }{2} } * \sigma }{E} ]^2$

=> $n =[ \frac{ 1.96 * 100}{ 0.02} ]^2$

=> $n = 96040000$

i $\to$ b

At $E_1 = 0.04$ and confidence level = 95% => $\alpha_1 = 0.05$ => $Z_{\frac{\alpha_1 }{2} } = 1.96$

$n_1 = [ \frac{Z_{\frac{\alpha_2 }{2} } * \sigma }{E_1} ]^2$

=> $n_1 =[ \frac{ 1.96 * 100}{ 0.04} ]^2$

=> $n_1 =24010000$

i $\to$ c

At $E_2 = 0.04$ confidence level = 99% => $\alpha_2 = 0.01$

The critical value of $\frac{\alpha_2 }{2}$ from the normal distribution table is

$Z_{\frac{ \alpha_2 }{2} } = 2.58$

=> $n_2 = [ \frac{Z_{\frac{\alpha_2 }{2} } * \sigma }{E_2} ]^2$

=> $n_2 =[ \frac{ 2.58 * 100}{ 0.04} ]^2$

=> $n_2 =41602500$

Considering ii

Given that the level of significance is $\alpha = 0.10$

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

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the margin of error is mathematically represented as

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

substituting values

$E = 1.645 * \frac{100 }{\sqrt{8} }$

$E = 58.16$

Generally the 90% confidence interval is mathematically evaluated as

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

=> $350 - 58.16 < \mu < 350 + 58.16$

=> $291.84 < \mu < 408.16$

So the interpretation is that there is 90% confidence that the mean fee charged to H&R Block customers last year is in the interval .So there is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval.

8 0

3 years ago

Evaluate 11+(−7*7)−9.

Butoxors [25]

Answer:

-47

Step-by-step explanation: Remember p.e.m.d.a.s

so start with (-7*7) = -49

then add so 11 + -49 = -38

then subtract so -38 - 9 = -47

6 0

3 years ago

Make a reasonable estimate for the situation. A snail moves at a speed of 3.395 inches per minute. If the snail keeps moving at

iren [92.7K]

Answer:

Distance = 21.0049 inches

Step-by-step explanation:

Given the following data;

Speed = 3.395 inches per minute.

Time = 6.187 minutes

To find the distance;

Distance = speed * time

Substituting into the equation, we have;

Distance = 3.395 * 6.187

Distance = 21.0049

4 0

3 years ago