T is a type of free verse poetry that has seventeen syllables (three lines of five seven, and five).

Fraction word problems Your strip of paper represents one lap on a track. Three students ran a relay and took turns running equa

artcher [175]

Answer:

A quarter $(\dfrac{1}{4})$ of one lap of the track.

Step-by-step explanation:

Three students ran ran a relay and took turns running equal parts of the track.

The race was three-fourths of a lap long.

Let the length of one lap of the track=x

The length of the race $=\frac{3}{4}x$

Since each of the students ran equal part,

Length run by each student

$=\dfrac{3}{4}x \div 3\\=\dfrac{3}{4}x X \dfrac{1}{3}\\=\dfrac{1}{4}x$

Therefore, each student ran a quarter $(\dfrac{1}{4})$ of one lap of the track.

5 0

3 years ago

Solve pls |2c+7|-3>4<br> :(

kotegsom [21]

Answer:

E

Step-by-step explanation:

2c + 7 > 0

2c > -7

c > -7/2

If c < -7/2

-2c -7 -3 > 4

-2c > 14

c < -7

if c > -7/2

2c + 7 -3 >4

2c >0

c >0

4 0

3 years ago

How can you find the area of an irregular polygon using area formulas?

algol [13]

Divide the polygon into separate shapes. Like a rectangle and then a triangle.
find the areas of those shapes and then add them together.

6 0

3 years ago

the table shows how the number of sit-ups Maria does each day has changed over time. At this rate, how many sit-ups will she do

tresset_1 [31]

Answer:

61

Step-by-step explanation:

a = 17

d = 21 - 17 = 4

12th day: 17 + 4(12 - 1)

17 + 44

61

8 0

3 years ago

Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro

Vinvika [58]

Answer:

40.1% probability that he will miss at least one of them

Step-by-step explanation:

For each target, there are only two possible outcomes. Either he hits it, or he does not. The probability of hitting a target is independent of other targets. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

0.95 probaiblity of hitting a target

This means that $p = 0.95$

10 targets

This means that $n = 10$

What is the probability that he will miss at least one of them?

Either he hits all the targets, or he misses at least one of them. The sum of the probabilities of these events is decimal 1. So

$P(X = 10) + P(X < 10) = 1$

We want P(X < 10). So

$P(X < 10) = 1 - P(X = 10)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{10,10}.(0.95)^{10}.(0.05)^{0} = 0.5987$

$P(X < 10) = 1 - P(X = 10) = 1 - 0.5987 = 0.401$

40.1% probability that he will miss at least one of them

7 0

3 years ago