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3 years ago

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A month of the year is chosen at random. What is the probability that the month starts with the letter J or the letter M?

2 answers:

Slav-nsk [51]3 years ago

4 0

5/12 because there is that about of months that either have J or M and it’s out of the amount of months there is

densk [106]3 years ago

3 0

Answer:

5/12

Step-by-step explanation:

You might be interested in

Answers to this question <br><br>8<br>2.5<br>12<br>3

ladessa [460]

Answer:

3 = ?

Step-by-step explanation:

The triangles are similar, so we can use ratios

15 12 -2

------ = --------------

15+? 12

15 10

------ = --------------

15+? 12

Using cross products

15 * 12 = 10 ( 15+?)

180 = 10 ( 15+?)

Divide each side by 10

18 = 15+?

Subtract 15 from each side

18-15 = 15+?-15

3 = ?

5 0

2 years ago

The length of a school bus is 12.6 meters. If 9 school buses park end to end with 2 meters between each one, what's the total le

weqwewe [10]

Answer:

129.4

Step-by-step explanation:

Calculation for what's the total length from the front of the first bus to the end of the last

First step is to multiply the given number of buses by the length of each one

Hence,

12.6 meters x 9 = 113.4 meters

Second step

Since there are 8 spaces in between we would multiply the 8 spaces by 2 meters

8 x 2 meters = 16

Now let calculate total length from the front of the first bus to the end of the last

16 + 113.4 meters =129.4

Therefore total length from the front of the first bus to the end of the last will be 129.4

7 0

3 years ago

One more question on this test guys. please answer as fast as possible. If you do,

GaryK [48]

1. The given rectangular equation is $x=2$ .

We substitute $x=r\cos \theta$ .

$r\cos \theta=2$

Divide through by $\cos \theta$

$r=\frac{2}{\cos \theta}$

$r=2}\sec \theta$

$\boxed{x=2\to r=2\sec \theta}$

2. The given rectangular equation is:

$x^2+y^2=36$

This is the same as:

$x^2+y^2=6^2$

We use the relation $r^2=x^2+y^2$

This implies that:

$r^2=6^2$

$\therefore r=6$

$\boxed{x^2+y^2=36\to r=6}$

3. The given rectangular equation is:

$x^2+y^2=2y$

This is the same as:

We use the relation $r^2=x^2+y^2$ and $y=r\sin \theta$

This implies that:

$r^2=2r\sin \theta$

Divide through by r

$r=2\sin \theta$

$\boxed{x^2+y^2=2y\to r=2\sin \theta}$

4. We have $x=\sqrt{3}y$

We substitute $y=r\sin \theta$ and $x=r\cos \theta$

$r\cos \theta=r\sin \theta\sqrt{3}$

This implies that;

$\tan \theta=\frac{\sqrt{3}}{3}$

$\theta=\frac{\pi}{6}$

$\boxed{x=\sqrt{3}y\to \theta=\frac{\pi}{6}}$

5. We have $x=y$

We substitute $y=r\sin \theta$ and $x=r\cos \theta$

$r\cos \theta=r\sin \theta$

This implies that;

$\tan \theta=1$

$\theta=\frac{\pi}{4}$

$\boxed{x=y\to \theta=\frac{\pi}{4}}$

6 0

3 years ago

What is the value of x after these statements if the starting value of x is 3.

kolbaska11 [484]

[A]
$3^{2}=9$
$5^{3}=125$
$9\neq25$

Ignore

[B]
$3^{2} = 9$
$9 \ \textgreater \ 3$
Yes

$3^{3}=27$
$4( 3^{2})=36

27 \ \textless \ 36$
Yes

Therefore
$x=3+2=5$

[C]
$4^{5}=1024

 5^{4}=625

1024\ \textless \ 625$
No

$5^{5}=3125

 5^{5}=3125

3125\ \textgreater \ 3125$
No

Ignore

[D]
$5^{5}=3125

 2^{5}=32

3125\ \textgreater \ 32$
Yes

$5^{2}=25
25\ \textless \ 11$
No

Therefore
$x=5+8=13$

[E]
$8\ \textgreater \ 7$
Yes

Therefore
$x = 13^{2}=169$

4 0

3 years ago

The ratio of the number of cookies to the number of candies in different gift boxes is 5:7. Which table best shows the number of

Margarita [4]

Answer: 10:14, 15:21, 20:28

Step-by-step explanation: These are all multiples of 5 and 7.

5: 10, 15, 20

7: 14, 21, 28

So it is C. 10:14, 15:21, 20:28

7 0

2 years ago

Read 2 more answers