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3 years ago

The price of a wallet is 31.95 and the sales tax is 7.5%

Mathematics

1 answer:

rewona [7]3 years ago

5 0

The sales tax is $2.40
So the final price is $34.35

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garri49 [273]

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8 0

2 years ago

Use the function g(x)-x+2 with initial values of 4, 2, and 1. What happens

kolezko [41]

Answer:

Step-by-step explanation:

The given function is , g(x)= - x + 2

For values of , is the sequence, -2,4,-2,4,-2,4........

For values of , is the sequence, 0,2,0,2........

For values of , is the sequence, 1,1,1,1,........

For initial value, of 4,the sequence converges to two values that is -2, 4.

For initial value of 2, the sequence again converges to two values which are 0,2.

For initial value of 1, the sequence converges to one value which is 1.

→→First Iterative value that is (-2)=Second iterative value (0) + 2 ×Third iterative value(1)

→→ First Iterative value that is (4)=Second iterative value (2) + 2 ×Third iterative value(1)

4 0

3 years ago

Factor 60x−84x using the gcf

ozzi

-24x
thatll be the answer for gcf

6 0

2 years ago

Rebecca had 9/10 of a yard of ribbon.Rebecca had 9/10 of a yard of ribbon. Then she gave 7/20 of a yard to a friend. How much do

scZoUnD [109]

I think it's 11/20.

6 0

3 years ago

A city hosted a music festival that included three concerts. According to the sales database, 28% of the audience attended the f

Vitek1552 [10]

Answer:

$P(X\cap Y\cap Z)=0.05$

Step-by-step explanation:

From the question we are told that:

Percentage of audience in first concert $P(X)=0.28$

Percentage of audience in second concert $P(Y)=0.42$

Percentage of audience in third concert $P(Z)=0.30$

Audience Percentage at at-least one concert $P(X \cup Y \cup Z)=0.80$

Percentage of audience at first & second concert $P(X \cap Y)=0.10$

Percentage of audience in first & third concert $P(X \cap Z)=0.08$

Percentage of audience in second & third concert $P(Y\cap Z)=0.07$

Generally the equation for probability of attending all concerts $P(X\cap Y\cap Z)$ is mathematically given by

$P(X \cup Y \cup Z)=P(X)+P(Y)+P(Z)-P(X \cap Y)-P(X \cap Z)-P(Y\cap Z)+P(X\cap Y\cap Z)$

$P(X\cap Y\cap Z)=P(X \cup Y \cup Z)-P(X)-P(Y)-P(Z)+P(X \cap Y)+P(X \cap Z)+P(Y\cap Z)$

$P(X\cap Y\cap Z)=0.80-0.28-0.42-0.30+0.10+0.80+0.70$

$P(X\cap Y\cap Z)=0.05$

Therefore the probability that a randomly selected audient attended all the concerts

$P(X\cap Y\cap Z)=0.05$

8 0

3 years ago