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3 years ago

I need a little help. (The question is on the screenshot by the way)

Mathematics

2 answers:

vagabundo [1.1K]3 years ago

7 0

Answer:

Step-by-step explanation:

the answer iud D

Olegator [25]3 years ago

5 0

Answer:

$\boxed{8} < \sqrt{80} < \boxed{9}$

Step-by-step explanation:

$\sqrt{80} =8.94427191$

Since, 8.94427191 falls between 8 and 9.

$\therefore \: \boxed{8} < \sqrt{80} < \boxed{9}$

You might be interested in

Two Social Security numbers (see Exercise 8.12) match zeros if a digit of one number is zero iff the corresponding digit of the

Roman55 [17]

Answer:

Proved

Step-by-step explanation:

From the given parameters, we have:

$n = 9$ i.e. the length of each security numbers

$r = 2$ i.e. 2 security numbers

Required

In 513 security numbers, 2 must have matching zeros

To do this, we make use of Pigeonhole principle.

First, we calculate the number of all security numbers not having matching zeros.

Each of the 9 digits can be selected in 2 ways.

2 ways implies that each digit is either 0 or not

So, total selection is:

$Total = 2^9$

$Total = 512$

Apply Pigeonhole principle

The principle states that: suppose there are n items in m containers, where $n>m$ , then there is at least one container that contains more than 1 item.

This means that if there are 512 security number without matching zeros, then there is 1 (i.e. 512 + 1) with matching zeros.

$512 + 1 = 513$

8 0

3 years ago

Question in pic! help asap!! will give brainliest

Viktor [21]

Answer:

a6b40c70

Step-by-step explanation:

a6b40c70

6 0

3 years ago

Read 2 more answers

At a two-hour basketball practice, Tracy can either choose to run miles, which take six minutes per mile, or play scrimmage game

Lera25 [3.4K]

Answer:

See the image attachment. It shows the three inequalities

$6m+30s \le 120$

$0 \le m \le 5$

$s \ge 0$

the inequalities are all stacked on each other with a large curly brace written off to the left

===========================================

Explanation:

Let,

m = number of miles Tracy runs

s = number of scrimmage games Tracy plays

m and s are integers

-------

Each mile takes 6 minutes, so running m miles takes up a total of 6m minutes.

Each scrimmage game takes 30 minutes, so in total she uses up 30s minutes of time if the number of games she plays is 's'.

In total, Tracy uses up 6m+30s minutes running and playing scrimmage games. The max time allotted for her is 2 hours or 120 minutes. Therefore, $6m+30s \le 120$ is one inequality that makes up the system of inequalities.

-------

We know that 'm' is positive, and we also know that m cannot get larger than 5, so we can write $0 \le m \le 5$ . This says "m is between 0 and 5 including both endpoints".

-------

For similar reasons in the last section above, we also know that 's' is positive as well; however we have no upper bound restrictions on 's'. So we simply say $s \ge 0$

-------

Overall, the system of inequalities has these three inequalities:

$6m+30s \le 120$

$0 \le m \le 5$

$s \ge 0$

they are all stacked on each other and written with a large curly brace to the left hand side of this stack. The large curly brace is just a way of keeping things together in a group or collection so to speak. A paired right curly brace is not included.

A visual reference of the answer is attached as an image below, just so you can see what I mean by the large curly brace.

3 0

4 years ago

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Anna [14]

Answer:

a) 0.4452

b) 0.0548

c) 0.0501

d) 0.9145

e) 6.08 minutes or greater

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.7 minutes

Standard Deviation, σ = 0.50 minutes.

We are given that the distribution of length of the calls is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(calls last between 4.7 and 5.5 minutes)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{4.7 - 4.7}{0.50} \leq z \leq \displaystyle\frac{5.5-4.7}{0.50}) = P(0 \leq z \leq 1.6)\\\\= P(z \leq 1.6) - P(z$

$P(4.7 \leq x \leq 5.5) = 44.52\%$

b) P(calls last more than 5.5 minutes)

$P(x > 5.5) = P(z > \displaystyle\frac{5.5-4.7}{0.50}) = P(z > 1.6)\\\\P( z > 1.6) = 1 - P(z \leq 1.6)$

Calculating the value from the standard normal table we have,

$1 - 0.9452 = 0.0548 = 5.48\%\\P( x > 5.5) = 5.48\%$

c) P( calls last between 5.5 and 6 minutes)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{5.5 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(1.6 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$

$P(5.5 \leq x \leq 6) = 5.01\%$

d) P( calls last between 4 and 6 minutes)

$P(4 \leq x \leq 6) = P(\displaystyle\frac{4 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(-1.4 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$