PLEASE ANSWERRRRRRRRR

On a bicycle ride,Betty rode 7 miles in 2 hours. At the same rate of speed,how far could she ride in 8 hours?Please hurry

prohojiy [21]

D 28 miles you want to multiply by 4 so 7x4 =28

3 0

3 years ago

Roger has two golden retrievers, sadie and buddy, buddy weighs 14 pounds more than sadie, if their total wight is 136 pounds, ho

Vlad1618 [11]

Answer:

A) buddy = sadie + 14

B) buddy + sadie = 136

A) buddy -sadie = 14

Adding equations A & B

2 * buddy = 150

buddy = 75 pounds

sadie = 61 pounds

Step-by-step explanation:

6 0

3 years ago

6th grade math<br><br> help on numbers 4 5 6 plz help

mario62 [17]

Answer:

4) 6 students

5) 7 fewer students

6) No, you can not tell from the frequency table how many students ran a mile in exactly 12 minutes.

Step-by-step explanation:

4) you’d look at the row from 8:00-8:59.

5) Add the first two rows together (6+2=8), then subtract that by the sum of the last two rows (9+6=15), which is 7

6) There’s no pattern in the frequency table, and the data points would be plotted differently since it’d be from a range of times, not one set time.

Hope this helped, sorry if I’m wrong on #6 ;)

6 0

3 years ago

Read 2 more answers

The sum of three consecutive odd integers is 69. find the integers

Marta_Voda [28]

Let the integers be x-2, x, x+2.

Given,

x -2 + x + x + 2 = 69

3x = 69

x = 23

Hence, the integers are 21, 23 and 25.

6 0

3 years ago

A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa

gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4 | red dice) = $\dfrac{1}{6}$

P (4 | green dice) = $\dfrac{3}{6}$ = $\dfrac{1}{2}$

A die is selected at random and rolled four times.

As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = $\dfrac{1}{2}$

The probability of two 1's and two 4's in the first dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^4$

= $\dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4$

= $6 \times ( \dfrac{1}{6})^4$

= $(\dfrac{1}{6})^3$

= $\dfrac{1}{216}$

The probability of two 1's and two 4's in the second dice can be calculated as:

= $\begin {pmatrix} \left \begin{array}{c}4\\2\\ \end{array} \right \end {pmatrix} \times \begin {pmatrix} \dfrac{1}{6} \end {pmatrix} ^2 \times \begin {pmatrix} \dfrac{3}{6} \end {pmatrix} ^2$

= $\dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $6 \times ( \dfrac{1}{6})^2 \times ( \dfrac{3}{6})^2$

= $( \dfrac{1}{6}) \times ( \dfrac{3}{6})^2$

= $\dfrac{9}{216}$

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = $\dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}$

The probability of two 1's and two 4's in both die = $\dfrac{1}{432} + \dfrac{1}{48}$

The probability of two 1's and two 4's in both die = $\dfrac{5}{216}$

By applying Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's ) = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's ) = $\dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}$

P(second die (green) | two 1's and two 4's ) = $\dfrac{0.5 \times 0.04166666667}{0.02314814815}$

P(second die (green) | two 1's and two 4's ) = 0.9

Thus; the probability the die chosen was green is 0.9

8 0

3 years ago