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2 years ago

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An insurance company uses the formula T=3.5n + 5 to work out the cost, T pounds, for insuring a customer for n days. Work out th

e cost for travel insurance for 4 days?

1 answer:

artcher [175]2 years ago

8 0

Answer:

Cost of travel insurance for 4 days = 19 pounds

Step-by-step explanation:

The insurance company uses the formula T = 3.5n + 5 to work out the cost , T pounds for insuring the customer for n number of days . Where n is the number of days.

The algebraic equation given has two variables which are cost T of insuring and n which is the number of days for insuring. Mathematically,

T = cost of insuring in pounds

n = number of days

Therefore, finding the cost T for 4 days

T = 3.5n + 5

T = 3.5 × 4 + 5

T = 14 + 5

T = 19 pounds

Cost of travel insurance for 4 days = 19 pounds

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A store owner wants to give a pen to each customer who visits their store. The table shows the cost for the number of pens in a

Flura [38]

Answer:

$9 for 36

Step-by-step explanation:

Divide the price by the quantity to receive the unit price. Here are the unit prices for each option:

1. $0.54

2. $0.43

3. $0.31

4. $0.28

5. $0.25

Basically, you want the lowest price. So, $9 for 36 is the best value because it costs $0.25 per pen.

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1 year ago

D2-9k+3 for d=10 and k=9

AVprozaik [17]

Answer:

22

Step-by-step explanation:

10^2-9(9)+3

100-81+3

19+3

22

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1 year ago

Willie and his brothers went to TGI Mondays. Their order consisted of 3 cheeseburgers at $8.75 each, a fish sandwich for $6.75,

Sladkaya [172]

<u>Answer:</u>$68.69

<u>Step-by-step explanation:</u>

Calculation of bill value

Cheeseburgers = 26.25

(8.75 x 3)

Fish Sandwich = 6.75

Pizza = 14.99

Sweet Tea

(4 x 2.09) = 8.36

Total of items = 56.35

Sales tax on food = 3.381

(56.35 x 6%)

Total bill value = 59.731

Calculation of tips

Bill value x 15%

=59.731 x 15%

Tips paid=8.95965

Calculation of meals cost

Total meal cost = Bill value + tips

=59.731+8.95965

=$68.69

Total meal cost is $68.69

7 0

1 year ago

4. Hydrocarbons in the cab of an automobile were measured during trips on the New Jersey Turnpike and trips through the Lincoln

pochemuha

Answer:

No these these result do not differ at 95% confidence level

Step-by-step explanation:

From the question we are told that

The first concentrations is $c _1= 30.0 \ g/m^3$

The second concentrations is $c _2 = 52.9 \ g/m^3$

The first sample size is $n_1 = 32$

The second sample size is $n_2 = 32$

The first standard deviation is $\sigma_1 = 30.0$

The first standard deviation is $\sigma_1 = 29.0$

The mean for Turnpike is $\= x _1 = \frac{c_1}{n} = \frac{31.4}{32} = 0.98125$

The mean for Tunnel is $\= x _2 = \frac{c_2}{n} = \frac{52.9}{32} = 1.6531$

The null hypothesis is $H_o : \mu _1 - \mu_2 = 0$

The alternative hypothesis is $H_a : \mu _1 - \mu_2 \ne 0$

Generally the test statistics is mathematically represented as

$t = \frac{\= x_1 - \= x_2}{ \sqrt{\frac{\sigma_1^2}{n_1} +\frac{\sigma_2^2}{n_2} }}$

$t = \frac{0.98125 - 1.6531}{ \sqrt{\frac{30^2}{32} +\frac{29^2}{32} }}$

$t = - 0.0899$

Generally the degree of freedom is mathematically represented as

$df = 32+ 32 - 2$

$df = 62$

The significance $\alpha$ is evaluated as

$\alpha = (C - 100 )\%$

=> $\alpha = (95 - 100 )\%$

=> $\alpha =0.05$

The critical value is evaluated as

$t_c = 2 * t_{0.05 , 62}$

From the student t- distribution table

$t_{0.05, 62} = 1.67$

So

$t_c = 2 * 1.67$

=> $t_c = 3.34$

given that

$t_c > t$ we fail to reject the null hypothesis so this mean that the result do not differ

6 0

1 year ago

Identify the relative maximum value of g(x) for the function shown below g(x)=7/x^2+5

Liula [17]

To find t<span>he relative maximum value of the function we need to find where the function has its first derivative equal to 0.

Its first derivative is -7*(2x)/(x^2+5)^2
</span>7*(2x)/(x^2+5)^2 =0 the numerator needs to be eqaul to 0
2x=0
x=0

g(0) = 7/5

The <span>relative maximum value is at the point (0, 7/5).</span>

8 0

1 year ago

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