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3 years ago

Nelson did 105 sit-ups in 3 minutes. What is the unit rate?

Mathematics

1 answer:

Oduvanchick [21]3 years ago

6 0

Answer:

The answer is 35

Step-by-step explanation:

You might be interested in

What was the equation of the graph below before it was shifted to the right 1.5 units?

Ira Lisetskai [31]

Answer:

B is correct.

{tex]G(x)=x^3-x[/tex]

Step-by-step explanation:

We are given a graph of cubic polynomial after shift 1.5 unit right.

Equation of given graph is

$G(x)=(x-1.5)^3-(x-1.5)$

For right and left shift, change in x value or horizontal translation.

If we shift "a" unit left then change in x

$x\rightarrow (x+a)$

If we shift "a" unit right then change in x

$x\rightarrow (x-a)$

Here, we need to shift 1.5 unit left to get original function.

$G(x)=(x-1.5)^3-(x-1.5)$

Shift 1.5 unit left

$F(x)=(x-1.5+1.5)^3-(x-1.5+1.5)$

$F(x)=x^3-x$

Hence, B is correct function.

6 0

3 years ago

Read 2 more answers

How many 2-digit numbers are multiples of 2 but not 3?

WARRIOR [948]

Answer:

30 numbers

Step-by-step explanation:

Two-digit numbers divisible by 2 are basically just the even two-digit numbers: 10, 12, 14,...96, 98.

Let's collectively divide all these numbers by 2:

10/2 , 12/2 , 14/2 , ... 96/2 , 98/2

5, 6, 7, ... 48, 49

Now subtract 4 from each of them:

5 - 4, 6 - 4, 7 - 4, ... , 48 - 4, 49 - 4

1, 2, 3, ... , 44, 45

So there are 45 multiples of 2 that are two-digit numbers.

The numbers that are both multiples of 2 and 3 are those that are multiples of 6:

12, 18, 24, ... , 96

We want to count how many there are, so divide all these numbers by 6:

12/6 , 18/6 , 24/6 , .... , 96/6

2, 3, 4, .... , 16

There are 15 numbers that are multiples of 2 and 3, so subtract 15 from 45: 45 - 15 = 30

The answer is 30.

Hope this helps!

7 0

4 years ago

An article in Technometrics (Vol. 19, 1977, p. 425) presents the following data on the motor fuel octane ratings of several blen

Slav-nsk [51]

Answer:

$Median = 90.4$

$Q_1 = 88.6$

$Q_3 = 92.2$

Step-by-step explanation:

Given

The above data

Required

- A stem and leaf display

- The median

- The quartiles

First, determine the range of the data

$Smallest = 83.4$

$Highest = 100.3$

Next, group each dataset base on common whole numbers.

So, we have:

$83.4$

$84.3\ 84.3$

$85.3$

$86.7\ 86.7\ 86.7$

$87.4\ 87.5\ 87.6\ 87.7\ 87.8\ 87.9$

$88.2\ 88.3\ 88.3\ 88.3\ 88.4\ 88.5\ 88.5\ 88.6\ 88.6\ 88.7\ 88.9$

$89.0\ 89.2\ 89.3\ 89.3\ 89.6\ 89.7\ 89.8\ 89.8\ 89.9\ 89.9$

$90.0\ 90.1\ 90.1\ 90.1\ 90.3\ 90.4\ 90.4\ 90.4\ 90.5\ 90.6\ 90.7\ 90.8\ 90.9$

$91.0\ 91.0\ 91.0\ 91.1\ 91.1\ 91.1\ 91.2\ 91.2\ 91.2\ \ 91.5\ 91.6\ 91.6\ 91.8\ 91.8$

$92.2\ 92.2\ 92.2\ 92.3\ 92.6\ 92.7\ 92.7\ 92.7$

$93.0\ 93.2\ 93.3\ 93.3\ 93.4\ 93.7$

$94.2\ 94.2\ 94.4\ 94.7$

$96.1\ 96.5$

$98.8$

$100.3$

Next, we construct the stem and leaf plot.

The whole numbers will be the stem while the decimal parts will be the leaf.

So, we have:

$\begin{array}{ccc}{Stem} & {} & {Leaf} & {83} & {|} & {.4} & {84} & {|} & {.3\ .3} & {85} & {|} & {.3} & {86} & {|} & {.7\ .7\ .7} & {87} &{|} & {.4\ .5\ .6\ .7\ .8\ .9} & {88} & {|} & {.2\ .3\ .3\ .3\ .4\ .5\ .5\ .6\ .6\ .7\ .9} &{89} & {|} & {.0\ .2\ .3\ .3\ .6\ .7\ .8\ .8\ .9\ .9} & {90} & {|} &{.0\ .1\ .1\ .1\ .3\ .4\ .4\ .4\ .5\ .6\ .7\ .8\ .9} & {91} &{|}&{.0\ .0\ .0\ .1\ .1\ .1\ .2\ .2\ .2\ .5\ .6\ .6\ .8\ .8} &{92} &{|} &{.2\ .2\ .2\ .3\ .6\ .7\ .7\ .7} \ \end{array}$

$\begin{array}{ccc} {93} & {|} & {.0\ .2\ .3\ .3\ .4\ .7} & {94} &{|} & {.2\ .2\ .4\ .7} &{96} & {|} & {.1\ .5} & {98} & {|} & {.8} & {100} &{|} &{.3} \ \end{array}$

From the above plot,

$n = 83$

The median is calculated as:

$Median = \frac{n+1}{2}th$

$Median = \frac{83+1}{2}th$

$Median = \frac{84}{2}th$

$Median = 42nd$

i.e. the median is at the 42nd position.

From the above stem and leaf plot.

The 42nd position is at stem 90 and the leaf .4

So the median is:

$Median = 90.4$

The lower quartile (Q) is calculated as:

$Q_1 = \frac{n+1}{4}th$

$Q_1 = \frac{83+1}{4}th$

$Q_1 = \frac{84}{4}th$

$Q_1 = 21st$

i.e. the lower quartile is at the 21st position.

From the above stem and leaf plot.

The 42nd position is at stem 88 and the leaf .6

So the lower quartile is:

$Q_1 = 88.6$

The upper quartile (Q3) is calculated as:

$Q_3 = 3 * \frac{n+1}{4}th$

$Q_3 = 3 * \frac{83+1}{4}th$

$Q_3 = 3 * \frac{84}{4}th$

$Q_3 = 3 * 21th$

$Q_3 = 63rd$

i.e. the upper quartile is at the 63rd position.

From the above stem and leaf plot.

The 63rd position is at stem 92 and the leaf .2

So the upper quartile is:

$Q_3 = 92.2$

8 0

3 years ago

Please help!

Bingel [31]

Answer: D

Step-by-step explanation:

5 0

3 years ago

According to the WHO MONICA Project the mean blood pressure for people in China is 128 mmHg with a standard deviation of 23 mmHg

Alja [10]

Answer:

a) Let X the random variable that represent the blood pressure for people of a population, and for this case we know the distribution for X is given by:

$X \sim N(128,23)$

Where $\mu=128$ and $\sigma=23$

b) $P(X\geq 135)=P(\frac{X-\mu}{\sigma}\geq \frac{135-\mu}{\sigma})=P(Z\geq \frac{135-128}{23})=P(Z\geq 0.304)$

And we can find this probability using the complement rule:

$P(Z\geq 0.304)=1-P(Z$

And in order to find this probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(Z\geq 0.304)=1-P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the blood pressure for people of a population, and for this case we know the distribution for X is given by:

$X \sim N(128,23)$

Where $\mu=128$ and $\sigma=23$

Part b

We are interested on this probability

$P(X\geq 135)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X\geq 135)=P(\frac{X-\mu}{\sigma}\geq \frac{135-\mu}{\sigma})=P(Z\geq \frac{135-128}{23})=P(Z\geq 0.304)$

And we can find this probability using the complement rule:

$P(Z\geq 0.304)=1-P(Z$

And in order to find this probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(Z\geq 0.304)=1-P(Z$

3 0

3 years ago