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3 years ago

Pls help i have been getting bad answers all day im about to fail 7th grade :( Pls I WILL GIVE BRAINLIEST ANYTHING :(

Mathematics

1 answer:

Vikentia [17]3 years ago

6 0

Answer:

1. a

2. c

3. math and the calculation of meat price per pound, then a comparison

Step-by-step explanation:

give brainliest pls pls pls

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Matthew biked for 4 days across an island. He started at the shoreline, which is at an elevation of 0 feet. At the end of day 1,

erica [24]

Answer:

The answer is 5,030 feet.

Step-by-step explanation:

It is stated that his starting point is at an elevation of 0 feet. He climbs 3.200 feet on the first day and then adds another 1.830 feet on the second day. This makes his total elevation 3,200 + 1,830 = 5.030 feet.

As he reaches the other side of the island on the 3rd day which is at an elevation of 0 feet as well, he loses all the elevation he gained on days 1 and 2, which is 5,030 feet.

I hope this answer helps.

3 0

3 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

4x^2-5x-15=0 how to find the vertex and y intercept

mezya [45]

$ax^2+bx+c=0\\\\the\ vertex:\left(\frac{-b}{2a};\frac{-(b^2-4ac)}{4a}\right)\\\\y-intrcept=c\\=======================================\\\\4x^2-5x-15=0\\\\a=4;\ b=-5;\ c=-15\\\\\frac{-b}{2a}=\frac{-(-5)}{2\cdot4}=\frac{5}{8};\ \frac{-(b^2-4ac)}{4a}=\frac{-[(-5)^2-4\cdot4\cdot(-15)]}{4\cdot4}=-\frac{265}{16}\\\\\boxed{the\ vertex:\left(\frac{5}{8};-\frac{265}{16}\right)}\\\\\boxed{y-intercept:-15}$