Help Please.. Write out mixed decimals using words.

p varies jointly as q and the square of r, and p = 200 when q = 2 and r = 3. Find p when q = 5 and r= 2.

Margarita [4]

Step-by-step explanation:

If a variables varies jointly, we can just divide it by the other variables in relation to it.

For example, since p variables jointly as q and square of r, then

$\frac{p}{q {r}^{2} } = k$

where k is a constant

First, let find k. Substitute p= 200

q= 2, and r=3.

$\frac{200}{2(3) {}^{2} } = k$

$\frac{200}{18} = k$

$\frac{100}{9} = k$

Now, since we know our constant, let find p.

$\frac{p}{q {r}^{2} } = \frac{100}{9}$

Q is 5, and r is 2.

$\frac{p}{5( {2}^{2}) } = \frac{100}{9}$

$\frac{p}{20} = \frac{100}{9}$

$p = \frac{2000}{9}$

3 0

2 years ago

I need to know the whole thing please help

aleksley [76]

Answer:

a) A student travelling to school on public transport: 15/52 or 0.231

b) A student walking to school: 16/52 or 0.308

c) A student not cycling to school: 43/52 or 0.827

Step-by-step explanation:

Total people = 52

Travel Method Frequency

Public Transport 12

Car 15

Cycle 9

Walk 16

Find the relative frequency of.

The formula used will be: $Relative\:Frequency=\frac{Given\: Frequency}{Size\: of\: sample \:space}$

a) A student travelling to school on public transport:

Given Frequency: 12

Size of sample space: 52

Apply formula: $Relative\:Frequency=\frac{Given\: Frequency}{Size\: of\: sample \:space}$

Fraction = 12/52

Decimal = 0.231

b) A student walking to school

Given Frequency: 16

Size of sample space. 52

Apply formula: $Relative\:Frequency=\frac{Given\: Frequency}{Size\: of\: sample \:space}$

Fraction = 16/52

Decimal = 0.308

c) A student not cycling to school.

We will consider all students except those who cycle.

12+15+16 = 43

Given Frequency: 43

Size of sample space. 52

Apply formula: $Relative\:Frequency=\frac{Given\: Frequency}{Size\: of\: sample \:space}$

Fraction = 43/52

Decimal = 0.827

6 0

3 years ago

If the weight (in grams) of cereal in a box of Lucky Charms is N(489,6), what is the probability that the box will contain less

Wittaler [7]

Answer:

0.000064 = 0.0064% probability that the box will contain less than the advertised weight of 466 g.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

N(489,6)

This means that $\mu = 489, \sigma = 6$