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2 years ago

Laquita practiced 60 math facts. Kinman practiced 15 more facts than Laquita. How many math facts did they practice in all?

Mathematics

1 answer:

Afina-wow [57]2 years ago

5 0

Answer:

They practiced a total of 75 math facts in all!

60 + 15 = 75 :D

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At a fast food burrito restaurant, customers ask servers to add particular ingredients to their burrito. One popular ingredient

Gnesinka [82]

25 ounces of guacamole are still expected to be available from batch.

Step-by-step explanation:

Given,

Quantity of guacamole in one serving = 2.2 ounces

Quantity of guacamole in one batch = 80 ounces

Quantity of guacamole in 25 servings = 2.2*25 = 55 ounces

Guacamole left = Quantity in batch - Quantity of 25 servings

Guacamole left = $80-55 = 25\ ounces\\$

25 ounces of guacamole are still expected to be available from batch.

Keywords: multiplication, subtraction

Learn more about subtraction at:

#LearnwithBrainly

5 0

3 years ago

I have no idea what to do with this someone help

joja [24]

The given distance of 5.20 would be A.

Replace A with 5.20 to solve for T.

T = 5.20^3/2
T = 11.9 years.

4 0

2 years ago

Read 2 more answers

What is the probability of choosing a card from a deck of cards that is a heart or a nine?

Mila [183]

8 out of 52. There are 4 nine cards and 4 heart cards. There are 52 cards in the standard deck. So it is an 8 out of 52 chance that you will choose either a heart or a nine.

3 0

2 years ago

Suppose that you pick a bit string from the set of all bit strings of length ten. Find the probability that a) the bit string ha

emmasim [6.3K]

Answer:

45/1024
1/4
15/128
193/512
9/512

Step-by-step explanation:

There are 2^10 = 1024 bit strings of length 10.

a) There are 10C2 = 45 ways to have exactly two 1-bits in 10 bits

p(2 1-bits) = 45/1024

b) Of the four (4) possibilities for beginning and ending bits (00, 01, 11, 10), exactly one (1) of those is 00.

p(b0=0 & b9=0) = 1/4

c) There are 10C7 = 120 ways to have seven 1-bits in the bit string.

p(7 1-bits) = 120/1024 = 15/128

d) ∑10Ck {for k=0 to 4} = 386 is the total of the number of ways to have 0, 1, 2, 3, or 4 1-bits in the string. If there are more than that, there won't be more 0-bits than 1-bits

p(more 0 bits) = 386/1024 = 193/512

e) The string will have two 1-bits if it starts with 1 and there is a single 1-bit among the other 9 bits. There are 9 ways that can happen, among the 512 ways to have 9 remaining bits.

p(2 1-bits | first is a 1-bit) = 9/512

6 0

2 years ago

A security expert claims that exactly 13% of all homeowners have a home security alarm. State H0 and Ha in words and in symbols

soldi70 [24.7K]

Answer:

Two tailed

Step-by-step explanation:

Given that a security expert claims that exactly 13% of all homeowners have a home security alarm.

Here we have to check whether the contention of security expert is correct or not

The test is to be done for the proportion of homeowners who have a home security alarm to total homeowners.

Let p be the proportion.

For this we set up hypotheses as:

$H_0 = p = 13%=0.13\\H_a: p \neq 0.13$

This is a two tailed test

Because alternate is for not equal to. If alternate is for > or < we say one tailed and if not equal to we say two tailed.

3 0

2 years ago