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Mnenie [13.5K]
3 years ago
7

Jason bought 7 new baseball trading cards to add to his collection. The next day his dog ate half of his collection. There are n

ow only 30 cards left. How many cards did he start with?
Mathematics
2 answers:
swat323 years ago
8 0

Answer:

53

Step-by-step explanation:

53+7=60-half=30

Stels [109]3 years ago
5 0

Answer:

geegegtkfifigjfjrjicirjjfigjrkivigk

Step-by-step explanation:

bgfeefcfifirifiridurjfjdjdjdjdj

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PLEASE HELP WITH #14 AND #7 PLEASE A.S.A.P !!!
kodGreya [7K]

Answer:

7.) 16.7

Step-by-step explanation:

7). First you have to find the length of the line with the triangle that is at a 90 angle with the longest side of the triangle

Using the pythagrorem theorem a² + b² = c²

4² + b² = 5²

b² = 5² - 4²

b² = 25 - 16

b² = 9

√b² =√9

b = 3

next you do the same thing to find the missing segment of the longest side

3² + b² = 13²

b² = 169 - 9

b² = 160

b = √160 or  ≈ 12.7

12.7 + 4 = 16.7

so 16.7 is the length of that side


3 0
4 years ago
What is the effective interest rate on an account that earns 5.5% compounded monthly?
nataly862011 [7]

Answer:

.46%

Step-by-step explanation:

.055/12= .00458= .46%

4 0
3 years ago
The side lengths of different triangles are given. Which triangle is a right triangle?
son4ous [18]

Answer:

D

Step-by-step explanation:

6 0
3 years ago
8,16,24,64 is the sequence
ra1l [238]
Can you restate the question so that I know what you are asking?:)
8 0
3 years ago
Three coins are tossed. Let the event H = all Heads and the event K = at least one Heads. (match like 1-a, etc)
Aleksandr [31]

Given:

Three coins are tossed.

Let the event H represents all Heads and the event K represents at least one Heads.

To find:

a. The probability that the outcome is all heads if at least one coin shows a heads.

b. P(K) = ?

c. P(H∩K) = ?

Solution:

If three coins are tossed, then the total possible outcomes are:

{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Total outcomes = 8

Possible outcomes for all Heads = 1

Possible outcomes for at least one Heads = 7

Let the following events:

H = all Heads

K = at least one Heads.

Then,

H=\{HHH\}

K=\{HHH, HHT, HTH, HTT, THH, THT, TTH\}

H\cap K=\{HHH\}

Now,

P(K)=\dfrac{7}{8}

P(H\cap K)=\dfrac{1}{8}

a. The probability that the outcome is all heads if at least one coin shows a heads is:

P(H|K)=\dfrac{P(H\cap K)}{P(K)}

P(H|K)=\dfrac{\dfrac{1}{8}}{\dfrac{7}{8}}

P(H|K)=\dfrac{1}{7}

Therefore, the probability that the outcome is all heads if at least one coin shows a heads is \dfrac{1}{7}.

b. P(K)=\dfrac{7}{8}

c. P(H\cap K)=\dfrac{1}{8}

8 0
3 years ago
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