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Solnce55 [7]
4 years ago
13

Please help me due tmm second question

Mathematics
1 answer:
AURORKA [14]4 years ago
6 0
Not sure if the math is correct but I think that it is $199.5
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What is the answer of the equation 7n+2= 4n+17
beks73 [17]

Answer:

n = 5

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

3 0
3 years ago
By what would you multiply each side of the equation ax=27 to solve x
olga nikolaevna [1]

Answer:

Step-by-step explanation:

ax= 27........I would normally solve this by division....but it says multiply....so multiply both sides by 1/a

example :

3x = 27.....multiply both sides by 1/3

1/3(3/1)x = 27 * 1/3

x = 27/3 which reduces to 9

u have to multiply by the reciprocal of a to get rid of a

3 0
3 years ago
Read 2 more answers
Select all the levels of measurement for which data can be qualitative.a. Ratiob. Intervalc. Ordinald. Nominal
Misha Larkins [42]

Answer:

C.) Ordinal

D.) Nominal

Step-by-step explanation: Qualitative variables simply refers to variables of non-numeric character, nature or property. They are usually employed in assigning variables into discreet groups. The nominal and ordinal scales are employed when dealing with qualititative or categorical variables, with nominal variables used when the classes or groups involved do not follow any definite order or rank. Example include classifying sex into (male or Female), and other categorical classification of discreet nature. The ordinal scale is usually employed when dealing with ranked classifications such as level of importance[very important, Important, not so important, unimportant] or grade[distinction, excellent, very good, good, fair, poor].

6 0
4 years ago
The number of research articles in a prominent journal written by researchers during 1983–2003 can be approximated by U(t) = 4.1
Nimfa-mama [501]

Answer:

Step-by-step explanation:

From the information given:

We are to find; an expression for the total no. of articles written since 1983

∴

The total no. of articles written since 1983  =\int \limts ^t_0 U(t) dt

= \int ^t_0 \limits \dfrac{4.1 e^{0.6\ t}}{0.2+e^{0.6t}} \  dt

=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 x} \bigg| \bigg]^t_0

=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 x} \bigg| - \mathtt{In} \bigg|0.2+1\bigg|\bigg]

=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | \dfrac{0.2 + e^{0.6 x}}{1.2} \bigg|\bigg]

=\dfrac{41}{6} \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-\dfrac{41}{6} \mathtt{ In} \bigg | 1.2\bigg |

=6.833\times \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-1.2458

=6.833\times \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-1.25 \ \mathbf{thousand \ articles}

Therefore, the total number of articles written since 1983 is  =6.833\times \mathtt{In} \bigg | {0.2 + e^{0.6 x} \bigg|-1.25 \ \mathbf{thousand \ articles}

b. To find how many articles were being written  from 1983 to 2003

i.e. t = 2003 - 1983 = 20

∴

Total articles written from 1983 to  2003 is =\int \limts ^{20}_0 U(t) dt

= \int ^{20}_0 \limits \dfrac{4.1 e^{0.6\ t}}{0.2+e^{0.6t}} \  dt

=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 x} \bigg| \bigg]^{20}_0

=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{0.6 *20} \bigg| - \mathtt{In} \bigg|0.2+e^{0.6*0}\bigg|\bigg]

=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{12} \bigg| - \mathtt{In} \bigg|0.2+e^0\bigg|\bigg]

=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{12} \bigg| - \mathtt{In} \bigg|0.2+1\bigg|\bigg]

=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | 0.2 + e^{12} \bigg| - \mathtt{In} \bigg|1.2\bigg|\bigg]

=\dfrac{4.1}{0.6}\bigg [ \mathtt{In} \bigg | \dfrac{0.2 + e^{12}}{1.2} \bigg|\bigg]

= 80.75 thousand articles

3 0
3 years ago
Hayley and alex are counting their pocket money. hayley has three times as much money as alex. together they have £20 in total.
kondaur [170]
Alex's money = x
Hayley's money = 3x (she has three times as much money as him)

20 = x + 3x
20 = 4x
/4      /4
5   =  x

So Alex's money = 5. Hayley's money is 3 times his, so Hayley's money is 15.


3 0
3 years ago
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