Tristan asked his coworkers about how much time they spent commuting each morning.

Find the next number in the sequence 2, -4, 8, -16, 32

VladimirAG [237]

Answer:

- 64

Step-by-step explanation:

The rule is to multiply by - 2 to get the next number. The numbers in the sequence are 2, - 4, 8, - 16, 32. The sequence shows that the next number is - 64

Hope This Helps :)

8 0

4 years ago

Read 2 more answers

A bag contains 50 lottery balls, numbered 1-50. A ball is chosen at random, then a month of the year is selected. What is the pr

Morgarella [4.7K]

Answer:

4/75

Step-by-step explanation:

We have two sample spaces

1.Lottery balls

2. Month of the year

For lottery balls the sample space is S= 50

Multiples of 3 between 1 and 50 are

(3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48)= 16 in numbers

Hence the probability of choosing a multiple of 3

Pr(multiple of 3) = 16/50= 8/25

Also the sample space for months of the year is S= 12

Two months starts with letter m, March and may

Pr(of months with m) = 2/12= 1/6

the probability of choosing a multiple of 3 and a month starting with the letter M= 8/25*1/6= 8/150= 4/75

6 0

4 years ago

Given: △ABC; AB=BC, m∠BDA = 60°, BD=4 cm, BD ⊥ BA . Find: DC, AC.

vekshin1

Answer:

DC = 10.93 cm , AC = 9.8 cm

Step-by-step explanation:

From trigonometry;

⇒ Tan 60 = AB/BD

⇒AB = BD Tan 60 ( where BD = 4 cm )

⇒ AB = 6.93 cm

Also, AB=BC , therefore;

⇒ BC = 6.93 cm

⇒ Cos 60 = BD/AD

⇒ AD = BD/ Cos 60 = 4/Cos 60

⇒ AD = 8 cm

From Pythagoras theorem;

⇒ $AC^{2}$ = $AB^{2}$ + $BC^{2}$ = $(6.93)^{2}$ + $(6.93)^{2}$

⇒ AC = $\sqrt{96.05}$ = 9.80 cm

⇒ DC = BD + BC = 4 + 6.93

⇒ DC = 10.93 cm

8 0

3 years ago

X-intercept and y-intercepts og y=x^2-3

Rufina [12.5K]

Answer:

Part 1) The y-intercept is the point $(0,-3)$

Part 2) The x-intercepts are the points $(-\sqrt{3},0)\ and\ (\sqrt{3},0)$

Step-by-step explanation:

we have

$y=x^{2}-3$

This is a vertical parabola open upward

step 1

Find the y-intercept

we know that

The y-intercept is the value of y when the value of x is equal to zero

so

For x=0

$y=(0)^{2}-3$

$y=-3$

The y-intercept is the point (0,-3)

step 2

Find the x-intercepts

we know that

The x-intercept is the value of x when the value of y is equal to zero

so

For y=0

$x^{2}-3=0$

solve for x

Adds 3 both sides

$x^{2}-3+3=0+3$

$x^{2}=3$

take square root both sides

$x=(+/-)\sqrt{3}$

so

The x-intercepts are the points $(-\sqrt{3},0)\ and\ (\sqrt{3},0)$

3 0

3 years ago

The values of 60, 62, and 84 were common to both samples. The three values are identified as outliers with respect to the age-gr

postnew [5]

Answer:

Only 60 is an identified outlier

Step-by-step explanation:

Given

Age group: 40 to 50

$Min = 60$ $Q_1 = 70$ $Median= 73$

$Q_3 = 76$ $Max = 85$

Conditions for outlier

$Value > Q_3 + 1.5 * IQR$

$Value < Q_1 - 1.5 * IQR$

See attachment for complete question

Required

Which of 60, 62 and 84 is an outlier of age grout 40 to 50 years

First, calculate the IQR of group 40 to 50.

This is calculated as:

$IQR = Q_3 - Q_1$

Where:

$Q_3 = 76$ and $Q_1 = 70$

So:

$IQR = 76 - 70$

$IQR = 6$

Next, is to test the outlier conditions on each value (i.e. 60, 62 and 84)

Testing 60

Condition 1

$Value > Q_3 + 1.5 * IQR$

$60 > 76 + 1.5 * 6$

$60 > 76 + 9$

$60 > 85$ --- False

Condition 2

$Value < Q_1 - 1.5 * IQR$

$60 < 70 - 1.5 * 6$

$60 < 70 - 9$

$60 < 61$ --- True

Because one of the conditions is true, then 60 is an outlier of group 40 - 50 years

Testing 62

Condition 1

$Value > Q_3 + 1.5 * IQR$

$62> 76 + 1.5 * 6$

$62 > 76 + 9$

$62 > 85$ --- False

Condition 2

$Value < Q_1 - 1.5 * IQR$

$62 < 70 - 1.5 * 6$

$62 < 70 - 9$

$62 < 61$ --- False

Because both conditions are false, then 62 is not an outlier of group 40 - 50 years

Testing 84

Condition 1

$Value > Q_3 + 1.5 * IQR$

$84> 76 + 1.5 * 6$

$84 > 76 + 9$

$84 > 85$ --- False

Condition 2

$Value < Q_1 - 1.5 * IQR$

$84 < 70 - 1.5 * 6$

$84 < 70 - 9$

$84 < 61$ --- False

Because both conditions are false, then 84 is not an outlier of group 40 - 50 years

5 0

3 years ago