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3 years ago

1. f(x) = ?? - 4x? - 7x + 10

Mathematics

1 answer:

RSB [31]3 years ago

4 0

Answer:

first answer choice

Step-by-step explanation:

did it on edge

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Joe buys 2 shirts for $18.50 each and a belt for $12.36. He pays tax at a rate of 5.5% on his purchase. How much tax does he pay

kirill115 [55]

Answer:37$ for the shirt Sales tax: $2.04

Cost/Price before ST: $37.00

Total Cost/Price including ST: $39.04

for the belt

Sales tax: $0.68

■ Cost/Price before ST: $12.36

■ Total Cost/Price including ST: $13.04

Step-by-step explanation:

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3 years ago

Read 2 more answers

For the past 40 days. Naomi has been recording the number of customers at her restaurant between 10:00 AM and 11:00 AM. During t

igor_vitrenko [27]

Answer:

The experimental probability there will be fewer than 20 customers on the forty-first day is 5/8, .625, 62.5%.

Step-by-step explanation

4 0

3 years ago

I need number 22 answered

pochemuha

Answer:

y = -7.5 I believe

Step-by-step explanation:

6 0

3 years ago

Please answer number 1.

Aneli [31]

Answer: The two specific conjectures are in the first image, and the two general conjectures are in the second image.

7 0

2 years ago

A commuter crosses one of three bridges, A, B, or C, to go home from work. The commuter crosses bridge A with probability 1/3, c

mihalych1998 [28]

Answer:

0.1333 = 13.33% probability that bridge B was used.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Arrives home by 6 pm

Event B: Bridge B used.

Probability of arriving home by 6 pm:

75% of 1/3(Bridge A)

60% of 1/6(Bridge B)

80% of 1/2(Bridge C)

$P(A) = 0.75*\frac{1}{3} + 0.6*\frac{1}{6} + 0.8*\frac{1}{2} = 0.75$

Probability of arriving home by 6 pm using Bridge B:

60% of 1/6. So

$P(A \cap B) = 0.6*\frac{1}{6} = 0.1$

Find the probability that bridge B was used.

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1}{0.75} = 0.1333$

0.1333 = 13.33% probability that bridge B was used.

8 0

3 years ago