19. Marianna wants to buy a new tennis racket that

Kaley's family drove to DisneyLand for spring break. Her mom and dad shared the driving duties for a total of 24 hours. Her mom

qaws [65]

Answer:

Mom drove for 18 hours.

Dad drove for 6 hours.

Step-by-step explanation:

Given that:

Number of hours for the driving duty = 24 hours

Speed at which the mom drove = 75 miles per hour

Speed at which the dad drove = 60 miles per hour

Total distance driven by both of them = 1710 miles

To find:

Number of hours for which each person drive for ?

Solution:

Let number of hours driven by dad = $x$ hours

Let number of hours driven by mom = (24 - $x$ ) hours

Let us have a look at the formula for Distance in terms of Speed and Time.

$Distance = Speed \times Time$

Distance traveled by Mom = $75 \times x =75x \ miles$

Distance traveled by dad = $60(24-x) =144-60x$ miles

As per question statement:

$75x+1440-60x=1710\\\Rightarrow 15x=1710-1440\\\Rightarrow 15x = 270\\\Rightarrow x = 18\ hours$

So, Mom drove for 18 hours.

Dad drove for 6 hours.

3 0

3 years ago

Your teacher gives you a graph of the parent function y = x^2 and the graph of another function. You notice that the new functio

Arlecino [84]

It’s either B or C if it’s being translated to the right it’s B but if it’s being translated to the left it’s C

7 0

3 years ago

Read 2 more answers

A hybrid car can go 141 miles on 3 gallons how much miles can it go on 4 gallons

SCORPION-xisa [38]

188

141 divided by 3 is 47 so 47 times 4 is 188

8 0

3 years ago

Prudence has 20 pairs of gold earrings and 4 pairs of silver earrings in a drawer in her jewelry box. If she randomly pulls six

Delicious77 [7]

There is a 0.833 chance that she pulls a golden earring

7 0

3 years ago

Teacher raises A school system employs teachers at

Cerrena [4.2K]

By adding a constant value to every salary amount, the measures of

central tendency are increased by the amount, while the measures of

dispersion, remains the same

The correct responses are;

(a) The shape of the data remains the same

(b) The mean and median are increased by $1,000

(c) The standard deviation and interquartile range remain the same

Reasons:

The given parameters are;

Present teachers salary = Between $38,000 and $70,000

Amount of raise given to every teacher = $1,000

Required:

Effect of the raise on the following characteristics of the data

(a) Effect on the shape of distribution

The outline shape of the distribution will the same but higher by $1,000

(b) The mean of the data is given as follows;

$\overline x = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i}$

Therefore, following an increase of $1,000, we have;

$\overline x_{New} = \dfrac{\sum (f_i \cdot (x_i + 1000))}{\sum f_i} = \dfrac{\sum (f_i \cdot x_i + f_i \cdot 1000))}{\sum f_i} = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i} + \dfrac{\sum (f_i \cdot 1000)}{\sum f_i}$

$\overline x_{New} = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i} + \dfrac{\sum (f_i \cdot 1000)}{\sum f_i} = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i} + 1000 = \overline x + 1000$

Therefore, the new mean, is equal to the initial mean increased by 1,000

Median;

Given that all salaries, $x_i$ , are increased by $1,000, the median salary, $x_{med}$ , is also increased by $1,000

Therefore;

The correct response is that the median is increased by $1,000

(c) The standard deviation, σ, is given by $\sigma =\sqrt{\dfrac{\sum \left (x_i-\overline x \right )^{2} }{n}}$ ;

Where;

n = The number of teaches;

Given that, we have both a salary, $x_i$ , and the mean, $\overline x$ , increased by $1,000, we can write;

$\sigma_{new} =\sqrt{\dfrac{\sum \left ((x_i + 1000) -(\overline x + 1000)\right )^{2} }{n}} = \sqrt{\dfrac{\sum \left (x_i + 1000 -\overline x - 1000\right )^{2} }{n}}$

$\sigma_{new} = \sqrt{\dfrac{\sum \left (x_i + 1000 -\overline x - 1000\right )^{2} }{n}} = \sqrt{\dfrac{\sum \left (x_i + 1000 - 1000 - \overline x\right )^{2} }{n}}$

$\sigma_{new} = \sqrt{\dfrac{\sum \left (x_i + 1000 - 1000 - \overline x\right )^{2} }{n}} =\sqrt{\dfrac{\sum \left (x_i-\overline x \right )^{2} }{n}} = \sigma$

Therefore;

$\sigma_{new} = \sigma$ ; The standard deviation stays the same

Interquartile range;

The interquartile range, IQR = Q₃ - Q₁

New interquartile range, IQR $_{new}$ = (Q₃ + 1000) - (Q₁ + 1000) = Q₃ - Q₁ = IQR

Therefore;

The interquartile range stays the same

Learn more here:

brainly.com/question/9995782

6 0

2 years ago