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3 years ago

15

The data plan for Shawn's phone has 32,000 megabytes of storage for photos. She wants to increase the amount of data by 4%. If e

ach photo uses 5 megabytes, will she have enough new memory for 200 additional photos? EXPLAIN

2 answers:

mixer [17]3 years ago

4 0

You need to do 32,000/100 which is 320 and then multiply this by 4. This is 1280. Now you need to divide 1280 by 5 which is 256.

This means that she can get 256 new photos, so yes, she can get 200 more photos.

Hope this helps x

Elis [28]3 years ago

4 0

Lo voy a escribir en español, no se sorry

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3 years ago

Read 2 more answers

If events A and B are mutually exclusive, P(A or B) = 0.5, and P(B) = 0.3; then what is P(A)?

11Alexandr11 [23.1K]

Answer:

$P(A) = 0.2$

Step-by-step explanation:

We have that:

$P(A) = P(a) + P(A \cap B)$

In which P(a) is the probability of a but not b and $P(A \cap B)$ is the probability of both A and B.

By the same logic, we have that:

$P(B) = P(b) + P(A \cap B)$

Also

P(A or B) is

$P(A \cup B) = P(a) + P(b) + P(A \cap B)$

Events A and B are mutually exclusive

This means that $P(A \cap B) = 0$

$P(B) = 0.3 = P(b)$

$P(A \cup B) = P(a) + P(b) + P(A \cap B)$

$0.5 = P(a) + 0.3$

$P(a) = 0.2$

$P(A) = P(a) + P(A \cap B) = 0.2 + 0 = 0.2$

4 0

4 years ago

A store sells 8 colors of balloons with at least 28 of each color. How many different combinations of 28 balloons can be chosen?

Len [333]

Answer:

(a) $Selection = 6724520$

(b) $At\ most\ 12 = 6553976$

(c) $At\ most\ 8 = 6066720$

(d) $At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

Step-by-step explanation:

Given

$Colors = 8$

$Balloons = 28$ --- at least

Solving (a): 28 combinations

From the question, we understand that; a combination of 28 is to be selected. Because the order is not important, we make use of combination.

Also, because repetition is allowed; different balloons of the same kind can be selected over and over again.

So:

$n => 28 + 8-1$ $= 35$

$r = 28$

$Selection = ^{35}^C_{28$

$Selection = \frac{35!}{(35 - 28)!28!}$

$Selection = \frac{35!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29*28!}{7!28!}$

$Selection = \frac{35*34*33*32*31*30*29}{7!}$

$Selection = \frac{35*34*33*32*31*30*29}{7*6*5*4*3*2*1}$

$Selection = \frac{33891580800}{5040}$

$Selection = 6724520$

Solving (b): At most 12 red balloons

First, we calculate the ways of selecting at least 13 balloons

Out of the 28 balloons, there are 15 balloons remaining (i.e. 28 - 13)

So:

$n => 15 + 8 -1 = 22$

$r = 15$

Selection of at least 13 =

$At\ least\ 13 = ^{22}C_{15}$

$At\ least\ 13 = \frac{22!}{(22-15)!15!}$

$At\ least\ 13 = \frac{22!}{7!15!}$

$At\ least\ 13 = 170544$

Ways of selecting at most 12 =

$At\ most\ 12 = Total - At\ least\ 13$ --- Complement rule

$At\ most\ 12 = 6724520- 170544$

$At\ most\ 12 = 6553976$

Solving (c): At most 8 blue balloons

First, we calculate the ways of selecting at least 9 balloons

Out of the 28 balloons, there are 19 balloons remaining (i.e. 28 - 9)

So:

$n => 19+ 8 -1 = 26$

$r = 19$

Selection of at least 9 =

$At\ least\ 9 = ^{26}C_{19}$

$At\ least\ 9 = \frac{26!}{(26-19)!19!}$

$At\ least\ 9 = \frac{26!}{7!19!}$

$At\ least\ 9 = 657800$

Ways of selecting at most 8 =

$At\ most\ 8 = Total - At\ least\ 9$ --- Complement rule

$At\ most\ 8 = 6724520- 657800$

$At\ most\ 8 = 6066720$

Solving (d): 12 red and 8 blue balloons

First, we calculate the ways for selecting 13 red balloons and 9 blue balloons

Out of the 28 balloons, there are 6 balloons remaining (i.e. 28 - 13 - 9)

So:

$n =6+6-1 = 11$

$r = 6$

Selection =

$^{11}C_6 = \frac{11!}{(11-6)!6!}$

$^{11}C_6 = \frac{11!}{5!6!}$

$^{11}C_6 = 462$

Using inclusion/exclusion rule of two sets:

$Selection = At\ most\ 12 + At\ most\ 8 - (12\ red\ and\ 8\ blue)$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 170544+ 657800- 462$

$Only\ 12\ red\ and\ only\ 8\ blue\ = 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = Total - Only\ 12\ red\ and\ only\ 8\ blue$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 6724520 - 827882$

$At\ most\ 12\ red\ and\ at\ most\ 8\ blue = 5896638$

3 0

3 years ago

Ice begins melting at 32f or at 0 C. Lead melts at 621.5 F. How many degrees fahrenheit above the melting point of ice does lead

lana [24]

The difference would be 589.5 degrees fahrenheit.

8 0

4 years ago