1 card holds 10 stickers, 59 stickers. How many cards are needed

#89 will give brainliest for best answer!

Lady_Fox [76]

Answer:

Median: 11

Range: 12

25th Percentile: 9

75th Percentile: 14

Interquartile Range: 5

Step-by-step explanation:

The median is the line in the 'middle' of the box. The range is the largest number minus the smallest number. 25th Percentile is the first line of the box, 75th is 3rd line. IQR is the 75th Percentile - 25th Percentile.

Hope this helps! :)

6 0

2 years ago

Why is circle 1 similar to circle 2?

earnstyle [38]

Answer:

Option A.

Step-by-step explanation:

we know that

The circle 1 and circle 2 have the same center

The ratio of its radius is equal to the scale factor

The scale factor is equal to

$\frac{4.8}{1.2}=4$

<em>That means</em>

The radius of circle 1 is 4 times the radius of circle 2

therefore

Circle 1 is a dilation of circle 2 with a scale factor of 4

6 0

3 years ago

Read 2 more answers

Waiting on the platform, a commuter hears an announcement that the train is running five minutes late. He assumes the arrival ti

natima [27]

Answer:

D. 91%

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Less than 15 minutes.

Event B: Less than 10 minutes.

We are given the following probability distribution:

$f(T = t) = \frac{3}{5}(\frac{5}{t})^4, t \geq 5$

Simplifying:

$f(T = t) = \frac{3*5^4}{5t^4} = \frac{375}{t^4}$

Probability of arriving in less than 15 minutes:

Integral of the distribution from 5 to 15. So

$P(A) = \int_{5}^{15} = \frac{375}{t^4}$

Integral of $\frac{1}{t^4} = t^{-4}$ is $\frac{t^{-3}}{-3} = -\frac{1}{3t^3}$

Then

$\int \frac{375}{t^4} dt = -\frac{125}{t^3}$

Applying the limits, by the Fundamental Theorem of Calculus:

At $t = 15$ , $f(15) = -\frac{125}{15^3} = -\frac{1}{27}$

At $t = 5$ , $f(5) = -\frac{125}{5^3} = -1$

Then

$P(A) = -\frac{1}{27} + 1 = -\frac{1}{27} + \frac{27}{27} = \frac{26}{27}$

Probability of arriving in less than 15 minutes and less than 10 minutes.

The intersection of these events is less than 10 minutes, so:

$P(B) = \int_{5}^{10} = \frac{375}{t^4}$

We already have the integral, so just apply the limits:

At $t = 10$ , $f(10) = -\frac{125}{10^3} = -\frac{1}{8}$

At $t = 5$ , $f(5) = -\frac{125}{5^3} = -1$

Then

$P(A \cap B) = -\frac{1}{8} + 1 = -\frac{1}{8} + \frac{8}{8} = \frac{7}{8}$

If given the train arrived in less than 15 minutes, what is the probability it arrived in less than 10 minutes?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{7}{8}}{\frac{26}{27}} = 0.9087$

Thus 90.87%, approximately 91%, and the correct answer is given by option D.

3 0

2 years ago

What shape has 6 sides?

zaharov [31]

A hexagon has 6 sides. Hope this helps!

4 0

2 years ago

The football team has 30 more members than the basketball teams. If each group had 10 more members, the ratio of their membershi

Zanzabum

<span>Let x be the number of members on the football team. Let y be the number of members on the basketball team. equation 1: y = x - 30 equation 2: (x + 10) / (y + 10) = 3 / 2 2 * (x+10) = 3 * (y+10) Let's replace y with x - 30. 2 * (x+10) = 3 * (x - 30 + 10) 2x + 20 = 3x - 60 x = 80 y = x - 30 = 80 - 30 = 50 The football team has 80 members. The basketball team has 50 members.</span>

3 0

3 years ago