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Bingel [31]
1 year ago
14

In which interval is the radical function f of x is equal to the square root of the quantity x squared plus 2 times x minus 15 e

nd quantity increasing?
[3, ∞)
(4, ∞)
[–5, 3]
(–∞, –5] ∪ [3, ∞)
Mathematics
1 answer:
sergij07 [2.7K]1 year ago
3 0

Using function concepts, it is found that it is increasing on the interval:

(–∞, –5] ∪ [3, ∞)

---------------

The function is given by:

f(x) = \sqrt{x^2 + 2x - 15}

The graph is given at the end of this question.

  • If the function is pointing upwards, it is increasing. Otherwise, it is decreasing.
  • In the graph, it can be seen that it is pointing upwards for x of -5 and less, or 3 and higher, thus, the interval is:

(–∞, –5] ∪ [3, ∞)

A similar problem is given at brainly.com/question/13539822

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In a study on the fertility of married women conducted by Martin O’Connell and Carolyn C. Rogers for the Census Bureau in 1979,
Fed [463]

Answer:

we cannot conclude hat the proportion of wives married less than two years who planned to have children is significantly higher than the proportion of wives married five years

Step-by-step explanation:

Given that in a study on the fertility of married women conducted by Martin O’Connell and Carolyn C. Rogers for the Census Bureau in 1979, two groups of childless wives aged 25 to 29 were selected at random, and each was asked if she eventually planned to have a child. One group was selected from among wives married less than two years and the other from among wives married five years.

Let X be the group married less than 2 years and Y less than 5 years

                         X        Y     Total

Sample size   300   300    600

Favouring       240   288    528

p                      0.8     0.96  0.88

H_0: p_x=p_y\\H_a: p_x>p_y

p difference = -0.16

Std error for difference = \sqrt{0.88*0.12/600} =0.01327

Test statistic = p difference/std error=-6.03

p value <0.000001

Since p is less than alpha 0.05 we cannot conclude hat the proportion of wives married less than two years who planned to have children is significantly higher than the proportion of wives married five years

4 0
2 years ago
A rectangular football field is 64 meters wide and 100 meters long. A player runs from one corner of the field in a diagonal lin
joja [24]
Let the distance be d

Using pythagorean theorem,

d^2 = 64^2 + 100^2
d^2 = 14096
d = 118.7
d ≈ 119 


The player ran for 119 meters
8 0
2 years ago
Read 2 more answers
A car insurance company suspects that the younger the driver is, the more reckless a driver he/she is. They take a survey and gr
erastovalidia [21]

Answer:

The confidence interval for the difference in proportions is

-0.028\leq p_1-p_2 \leq 0.096

No. As the 95% CI include both negative and positive values, no proportion is significantly different from the other to conclude there is a difference between them.

Step-by-step explanation:

We have to construct a confidence interval for the difference of proportions.

The difference in the sample proportions is:

p_1-p_2=x_1/n_1-x_2/n_2=(183/217)-(322/398)=0.843-0.809\\\\p_1-p_2=0.034

The estimated standard error is:

\sigma_{p_1-p_2}=\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2} } \\\\\sigma_{p_1-p_2}=\sqrt{\frac{0.843*0.157}{217}+\frac{0.809*0.191}{398} } \\\\\sigma_{p_1-p_2}=\sqrt{0.000609912+0.000388239}=\sqrt{0.000998151} \\\\ \sigma_{p_1-p_2}=0.0316

The z-value for a 95% confidence interval is z=1.96.

Then, the lower and upper bounds are:

LL=(p_1-p_2)-z*\sigma_p=0.034-1.96*0.0316=0.034-0.062=-0.028\\\\\\UL=(p_1-p_2)+z*\sigma_p=0.034+1.96*0.0316=0.034+0.062=0.096

The confidence interval for the difference in proportions is

-0.028\leq p_1-p_2 \leq 0.096

<em>Can it be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group?</em>

No. It can not be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group, as the confidence interval include both positive and negative values.

This means that we are not confident that the actual difference of proportions is positive or negative. No proportion is significantly different from the other to conclude there is a difference.

8 0
2 years ago
A machine fills boxes weighing Y lb with X lb of salt, where X and Y are normal with mean 100 lb and 5 lb and standard deviation
Bas_tet [7]

Answer:

Option b. None is the correct option.

The Answer is 63%

Step-by-step explanation:

To solve for this question, we would be using the z score formula

The formula for calculating a z-score is given as:

z = (x-μ)/σ,

where

x is the raw score

μ is the population mean

σ is the population standard deviation.

We have boxes X and Y. So we will be combining both boxes

Mean of X = 100 lb

Mean of Y = 5 lb

Total mean = 100 + 5 = 105lb

Standard deviation for X = 1 lb

Standard deviation for Y = 0.5 lb

Remember Variance = Standard deviation ²

Variance for X = 1lb² = 1

Variance for Y = 0.5² = 0.25

Total variance = 1 + 0.25 = 1.25

Total standard deviation = √Total variance

= √1.25

Solving our question, we were asked to find the percent of filled boxes weighing between 104 lb and 106 lb are to be expected. Hence,

For 104lb

z = (x-μ)/σ,

z = 104 - 105 / √25

z = -0.89443

Using z score table ,

P( x = z)

P ( x = 104) = P( z = -0.89443) = 0.18555

For 1061b

z = (x-μ)/σ,

z = 106 - 105 / √25

z = 0.89443

Using z score table ,

P( x = z)

P ( x = 106) = P( z = 0.89443) = 0.81445

P(104 ≤ Z ≤ 106) = 0.81445 - 0.18555

= 0.6289

Converting to percentage, we have :

0.6289 × 100 = 62.89%

Approximately = 63 %

Therefore, the percent of filled boxes weighing between 104 lb and 106 lb that are to be expected is 63%

Since there is no 63% in the option, the correct answer is Option b. None.

3 0
2 years ago
Round 720,051,852 to the nearest hundred
marysya [2.9K]

Answer:

720,051,900

Step-by-step explanation:

it rounds up to 720,051,900 because 5 tells you to round up

4 0
1 year ago
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