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2 years ago

13

You and your friend are using ribbon to make cards. Your ribbon is 78 inch wide. Your friend's ribbon is 58 inch wide. How much

wider is the ribbon you are using?

2 answers:

Yakvenalex [24]2 years ago

4 0

Answer:

20 inches longer

Step-by-step explanation:

78-58= 20

serious [3.7K]2 years ago

4 0

20 bro come one that kindergarten math

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Lana71 [14]

Answer:

c°=70°

Step-by-step explanation:

taking that it would be an alternate interior totaling 180°

take the difference: 180°-110°=70°

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3 years ago

A football team has a squad of 24 players,

lora16 [44]

Answer: 1/3

Explanation:

1/12 of the squad are suspended:

1/12 x 24 = 24/12 = 2

So 2 player are suspended

24 - 2 = 22 player

And some are injured leaving 14 left to play:

22 - 14 = 8 players injured

So 8/24 is the fraction of injured player.

But 8/24 = 1/3 so we can say 1/3 of the whole squad are injured.

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3 years ago

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Answer: the function is monotonically increasing since the output values are continually getting larger. This also tells us that the end behavior of the function is infinity.

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Step-by-step explanation:

Plato answer

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3 years ago

Help!! A, B, C, or D

Elina [12.6K]

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3 years ago

67% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 48 owned dogs are r

oksian1 [2.3K]

Answer:

a) Probability that exactly 29 of them are spayed or neutered = 0.074

b) Probability that at most 33 of them are spayed or neutered = 0.66

c) Probability that at least 30 of them are spayed or neutered = 0.79

d) Probability that between 28 and 33 (including 28 and 33) of them are spayed or neutered = 0.574

Step-by-step explanation:

This is a binomial distribution question

probability of having a spayed or neutered dog, p = 0.67

probability of having a dog that is not spayed or neutered, q = 1 - 0.67

q = 0.23

sample size, n = 48

According to binomial distribution formula:

$P(X=r) = nCr p^r q^{n-r}$

where $nCr = \frac{n!}{(n-r)! r!}$

a) Probability that exactly 29 of them are spayed or neutered

$P(X= 29) = 48C29 * 0.67^{29} * 0.23^{19}\\P(X=29) = 0.074$

b) Probability that at most 33 of them are spayed or neutered

$P(X \leq 33) =1 - P(X > 33)\\P(X \leq 33) =1 - 0.34\\P(X \leq 33) = 0.66$

c) Probability that at least 30 of them are spayed or neutered

$P(X \geq 30) = 1 - P(x < 30)\\P(X \geq 30) = 1 - 0.21\\P(X \geq 30) = 0.79$

d) Probability that between 28 and 33 (including 28 and 33) of them are spayed or neutered.

$P(28 \leq X \leq 33) = P(X=28) + P(X=29) + P(X=30) + P(X=31) + P(X=32) + P(X=33)\\P(28 \leq X \leq 33) = 0.053 + 0.074 + 0.095 + 0.112 + 0.121 + 0.119\\P(28 \leq X \leq 33) = 0.574$

8 0

3 years ago