All categories

2 years ago

Need Help. The question is down below

Mathematics

1 answer:

madreJ [45]2 years ago

4 0

Useful Log Rules:

log(A*B) = log(A)+log(B) .......... log rule 1
log(A/B) = log(A) - log(B) .......... log rule 2
log(A^B) = B*log(A) ................... log rule 3

=========================================================

Part (a)

All logs shown below are base 3.

log(500) = log(5*100)

log(500) = log(5*10^2)

log(500) = log(5)+log(10^2) .... use log rule 1

log(500) = log(5) + 2*log(10) .... use log rule 3

log(500) = 1.4650 + 2*2.096 ....... substitution

log(500) = 5.657

<h3>Answer: 5.657</h3>

=========================================================

Part (b)

All logs shown below are base 3.

log(2) = log(10/5)

log(2) = log(10) - log(5) .... use log rule 2

log(2) = 2.096 - 1.4650 ....... substitution

log(2) = 0.631

<h3>Answer: 0.631</h3>

You might be interested in

PLEASE HELP! I HAVE TWO QUESTIONS AND WILL GIVE 100PTS AND BRAINLIEST IF CORRECT!

Volgvan

Answer:

1) $283.50

2) 65.21 meters.

Step-by-step explanation:

Part 1)

We know that a spool of ribbon holds 6.75 meters.

A craft club buys 21 spools.

So, the total length of ribbon bought is:

$T=21(6.75)=141.75\text{ meters}$

Each meter costs $2. So, the total cost will be:

$141.75(2)=\$ 283.50$

Part 2)

A spool holds 6.75 meters.

We bought 21 of them, so that total length we have is 141.75 meters.

The club used 76.54 meters.

So, the amount of ribbon that is left is represented by:

$141.75-76.54$

Subtract:

$=65.21\text{ meters}$

So, 65.21 meters of ribbon will be left.

7 0

2 years ago

Read 2 more answers

Samples of emissions from three suppliers are classified for conformance to air-quality specifications. The results from 100 sam

tigry1 [53]

Answer:

$P(A) = \frac{30}{100}$

$P(B) = \frac{77}{100}$

$P(A\ n\ B) = \frac{22}{100}$

$P(A\ u\ B) = \frac{85}{100}$

Step-by-step explanation:

Given

See attachment for proper format of table

$n = 100$ --- Sample

A = Supplier 1

B = Conforms to specification

Solving (a): P(A)

Here, we only consider data in sample 1 row.

In this row:

$Yes = 22$ and $No = 8$

So, we have:

$n(A) = Yes + No$

$n(A) = 22 + 8$

$n(A) = 30$

P(A) is then calculated as:

$P(A) = \frac{n(A)}{Sample}$

$P(A) = \frac{30}{100}$

Solving (b): P(B)

Here, we only consider data in the Yes column.

In this column:

$(1) = 22$ $(2) = 25$ and $(3) = 30$

So, we have:

$n(B) = (1) + (2) + (3)$

$n(B) = 22 + 25 + 30$

$n(B) = 77$

P(B) is then calculated as:

$P(B) = \frac{n(B)}{Sample}$

$P(B) = \frac{77}{100}$

Solving (c): P(A n B)

Here, we only consider the similar cell in the yes column and sample 1 row.

This cell is: [Supplier 1][Yes]

And it is represented with; n(A n B)

So, we have:

$n(A\ n\ B) = 22$

The probability is then calculated as:

$P(A\ n\ B) = \frac{n(A\ n\ B)}{Sample}$

$P(A\ n\ B) = \frac{22}{100}$

Solving (d): P(A u B)

This is calculated as:

$P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B)$

This gives:

$P(A\ u\ B) = \frac{30}{100} + \frac{77}{100} - \frac{22}{100}$

Take LCM

$P(A\ u\ B) = \frac{30+77-22}{100}$

$P(A\ u\ B) = \frac{85}{100}$

8 0

2 years ago

How to find the area only of a triangle

Cloud [144]

Area= height(base)/2

7 0

2 years ago

Tommy is trying to find the lengthy of one side of a square patio with area 289 feet squared what is the lengthy of one side

mars1129 [50]

Area of a square = L * L
As you know 17 * 17 = 289
So length of one side of square equals 17

6 0

3 years ago

The following table lists the masses, in grams, of the nine planets (including dwarf planet Pluto). Use the data to choose the c

Marrrta [24]

The ratio between the masses of Jupiter and Venus is

$\dfrac{1.989\cdot 10^{30}}{4.869\cdot 10^{27}}$

We can separate the coefficient and the powers of ten:

$\dfrac{1.989\cdot 10^{30}}{4.869\cdot 10^{27}}=\dfrac{1.989}{4.869}\cdot \dfrac{10^{30}}{10^{27}}$

Use the exponent rule

$\dfrac{a^b}{a^c}=a^{b-c}$

to simplify the powers of 10:

$\dfrac{10^{30}}{10^{27}}=10^3$

So, the ratio is

$\dfrac{1.989}{4.869}\cdot 10^3\approx 0.4\cdot 10^3=400$

So, Jupiter has a mass that is approximately 400 times greater than Venus's.

4 0

2 years ago

Read 2 more answers