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2 years ago

Can someone please help

Mathematics

1 answer:

Alika [10]2 years ago

3 0

Answer:

option (B).

Step-by-step explanation:

it's really simple....

just replace the (n) near the 3 with 1,2,3,4,5 and 6 and then solve the maths and when you're done , you add the answers all up.

sooo...

3(1)-2 = 1

3(2)-2 = 4

3(3)-2 = 7

3(4)-2 = 10

3(5)-2 = 13

3(6)-2 = 16

now we add them all up...

1+4+7+10+13+16 = 51.

Hope this helps.

Good luck.

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yaroslaw [1]

Answer:

∠TUY = 120°

Step-by-step explanation:

There are two rules you can apply here:

1) The sum of the angles inside a triangle is 180°. This means that ∠SUT = 180° - 70° - 50 ° = 60°

2) A vector coming out of a line two angles between itself and the line, and they add up to 180° as well. This means that ∠TUY = 180° - ∠SUT. So ∠TUY = 180° - 60° = 120°

So the missing angle one hundred and twenty degrees.

4 0

3 years ago

An economist wondered if people who go grocery shopping on weekdays go more or less often on Fridays than any other day. She fig

Kamila [148]

Answer:

There is not enough statistical evidence to state that the true proportion of grocery shoppers from Monday to Friday that goes most frequently on Fridays is different from 20%. The p - value is 0.0130.

Step-by-step explanation:

To solve this problem, we run a hypothesis test about the population proportion.

Proportion in the null hypothesis (p0) = 0.2

Sample size (n) = 75

Sample proportion (sp) = 24/75 = 0.32

Significance level = 0.01

H0: p = 0.2

Ha: p $\neq$ 0.2

Test statistic = (sp - p0) / sqrt ((p0) (1-p0) / n)

Left critical Z value (for 0.01) = -2.5758

Right critical Z value (for 0.01) = 2.5758

Calculated statistic = (0.32 - 0.2) / sqrt ((0.32) (1-0.32) / 75) = 2.2278

Since, -2.5758 < Test statistic < 2.5758, the null hypothesis cannot be rejected. There is not enough statistical evidence to state that the true proportion of grocery shoppers from Monday to Friday that goes most frequently on Fridays is different from 20%. The p - value is 0.0130.

5 0

3 years ago

Suppose a movie starts at 5:00 p.m. and Lindsay, a customer who is always late, arrives at the movie theater at a random time be

Arte-miy333 [17]

Answer:

a) We know that the distribution for X is given by: $X \sim Unif (a=10, b=45)$

The density function for this distirbution is given by:

$f(x) = \frac{1}{b-a}= \frac{1}{45-10}= \frac{1}{35} , 10 \leq X \leq 45$

So then the height of the uniform density curve would be given by:

$h = \frac{1}{35}= 0.029$

b) For this case we want to calculate the following probability:

$P(X\leq 20)$

And we can use the cumulative distribution function given by:

$F(X) =\frac{x-a}{b-a}=\frac{x-10}{35} , 10 \leq X \leq 45$

And we can calculate the probability like this:

$P(X \leq 20) = F(20) = \frac{20-10}{35}=0.286$

Step-by-step explanation:

For this case we define our random variable X="Lindsay's late arrival time, in minutes"

Part a

We know that the distribution for X is given by: $X \sim Unif (a=10, b=45)$

The density function for this distirbution is given by:

$f(x) = \frac{1}{b-a}= \frac{1}{45-10}= \frac{1}{35} , 10 \leq X \leq 45$

So then the height of the uniform density curve would be given by:

$h = \frac{1}{35}= 0.029$

Part b

For this case we want to calculate the following probability:

$P(X\leq 20)$

And we can use the cumulative distribution function given by:

$F(X) =\frac{x-a}{b-a}=\frac{x-10}{35} , 10 \leq X \leq 45$

And we can calculate the probability like this:

$P(X \leq 20) = F(20) = \frac{20-10}{35}=0.286$

7 0

4 years ago

Tony's cat weighs c pounds. Three months from know, the cat's weight is expected to increase by 12.5% Write two expressions to r

horrorfan [7]

Answer:

Step-by-step explanation:

Tony's cat weighs c pounds. This means that the cat currently weighs c pounds.

Three months from know, the cat's weight is expected to increase by 12.5%. This means that in three months time, the weight of the cat is expected to be the sum of its current rate × 12.5 % of its current rate. It becomes

c + 12.5/100 × c = c + 0.125c

= 1.125c

So the two expressions that represent the cat's weight in three months would be

1) three months from now, the cat's weight will increase by 0.125c

2) three months from now, the new weight of the cat will be 1.125c

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4 years ago

Someone please help fast it’s for a test

Vesnalui [34]

Answer:

The answer is 15

Step-by-step explanation: They give you the answer for x, f(x) = f(2)

f(2)=2(2)^2 + 6(2) - 5 f(2)=2(4) + 6(2) - 5 f(2)=8 + 12 - 5 f(2)=15

6 0

3 years ago

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