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3 years ago

Joe ran 3 miles yesterday and wants to run at least 15 miles

Mathematics

1 answer:

igomit [66]3 years ago

4 0

Answer: 3(x)≥15 →x≥5

Step-by-step explanation:

Set it up with x as the number of days. Then multiply that by 3, so that you get 3(x). This will show that for an x amount of days, you ran 3 miles. then set it to MORE THAN or EQUAL TO 15. Then divide both sides by 3, and you get the above answer.

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Write down your full calculator display.

Alja [10]

Answer:

83.615 515 710 687 7

Step-by-step explanation:

see the picture I gave

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4 years ago

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Color blindness is an inherited characteristic that is more common in males than females. Let F represent female, and C represen

malfutka [58]

Answer:

0.513 = 51.3% probability of being a female.

Step-by-step explanation:

We are given the following formula:

$P(F \cup C) = P(F) + P(C) - P(F \cap C)$

In which:

Event F: Female

Event C: Red-green color blindness.

Probability of a Female

We have to find P(F), which is given by:

$P(F) = P(F \cup C) + P(F \cap C) - P(C)$

We are given by:

$P(C) = 0.039, P(F \cap C) = 0.004, P(F \cup C) = 0.548$

$P(F) = P(F \cup C) + P(F \cap C) - P(C) = 0.548 + 0.004 - 0.039 = 0.513$

0.513 = 51.3% probability of being a female.

5 0

3 years ago

Eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

LenaWriter [7]

Answer:

3/4

Step-by-step explanation:

75%= 75/100

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3 years ago

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A magazine conducts an annual survey in which readers rate their favorite cruise ship. All ships are rated on a 100-point scale,

Lyrx [107]

Answer:

a) The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =85.82-81.90=3.92$

b) $SE=\sqrt{(\frac{4.55^2}{35}+\frac{3.97^2}{44})}=0.975$

And the Margin of error is given by:

$Me= z_{\alpha/2} * SE=1.96*0.975=1.910$

c) The 95% confidence interval would be given by $2.009 \leq \mu_1 -\mu_2 \leq 5.83$ .

Step-by-step explanation:

Notation and previous concepts

$n_1 =35$ represent the sample of ships that carry fewer than 500 passengers

$n_2 =44$ represent the sample of ships that carry 500 or more passengers

$\bar x_1 =85.82$ represent the mean sample of of ships that carry fewer than 500 passengers

$\bar x_2 =81.90$ represent the mean sample of of ships that carry 500 or more passengers

$\sigma_1 =4.55$ represent the population deviation of ships that carry fewer than 500 passengers

$\sigma_2 =3.97$ represent the sample deviation of ships that carry 500 or more passengers

$\alpha=0.05$ represent the significance level

Confidence =95% or 0.95

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{(\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2})}$ (1)

Part a

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =85.82-81.90=3.92$

Part b: At 95% confidence, what is the margin of error?

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$