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3 years ago

I need the answer to this ASAP

Mathematics

1 answer:

scoray [572]3 years ago

5 0

Answer: B. NO

Step-by-step explanation: It does not pass the parallel line test

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I need help with this question

tia_tia [17]

26.5 goes in the first box
The answer is 689.

8 0

2 years ago

Read 2 more answers

Of the 500 cars displayed in a certain car dealer, 7 percent are blue and 4 percent are red. The number of blue cars in the car

solong [7]

Answer:

70 percent

Step-by-step explanation:

7 0

2 years ago

Suppose the scores on a test given to all juniors in a school district are normally distributed with a mean of 74 and a standard

ivann1987 [24]

Answer:

$\mu -\sigma = 66 , \mu +\sigma = 82$

$\mu -2\sigma = 58 , \mu +2\sigma = 90$

$\mu -3\sigma = 50 , \mu +3\sigma = 98$

And in the figure attached we see the limits with the percentages associated.

Step-by-step explanation:

For this case we know that the random variable of interest is the scores on a test given to all juniors in a school district follows a normal distribution with the following parameters:

$X \sim N(\mu= 74, \sigma =8)$

For this case we know from the empirical rule that within one deviation from the mean we have approximately 68.2% of the data, within 2 deviations from the mean we have 95% and within 3 deviation 99.7%

We can find the limits and we got:

$\mu -\sigma = 66 , \mu +\sigma = 82$

$\mu -2\sigma = 58 , \mu +2\sigma = 90$

$\mu -3\sigma = 50 , \mu +3\sigma = 98$

And in the figure attached we see the limits with the percentages associated.

7 0

3 years ago

A marathon runner runs 7.6 mph for 3 and half hours how many miles has he run

igomit [66]

7.6 x 3.5 = 26.6 miles

6 0

3 years ago

Two cards are drawn in succession from a standard 52-card deck.1. What is the probability that the first card is red and the sec

rodikova [14]

Answer:

52 cards:

26 red and 26 black

P(R) = probability of picking a red card

P(B) = probability of picking a black card

P(R) = P(B) = ¹/₂

If with replacement:

P(R∩B) = (¹/₂)(¹/₂) = ¹/₄

If without replacement:

P(R∩B) = (¹/₂)(²⁶/₅₁) = ¹³/₅₁

8 Balls:

3 red and 5 white

P(R) = probability of picking a red ball

P(W) = probability of picking a white ball

P(R) = ³/₈

P(W) = ⁵/₈

If with replacement:

P(R∩W) ∪ P(W∩R) = (³/₈)(⁵/₈) + (⁵/₈)(³/₈)

= ¹⁵/₆₄ + ¹⁵/₆₄

= ³⁰/₆₄

= ¹⁵/₃₂

If without replacement:

P(R∩W) ∪ P(W∩R) = (³/₈)(⁵/₇) + (⁵/₈)(³/₇)

= ¹⁵/₄₂ + ¹⁵/₄₂

= ³⁰/₄₂

= ⁵/₇

8 0

2 years ago