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2 years ago

Can someone help me on this

Mathematics

1 answer:

PtichkaEL [24]2 years ago

6 0

Don’t you just add them? So it is 200?

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Evaluate t-7 for t =20

Luda [366]

Answer:

The answer is 14

Step-by-step explanation:

t - 7 ; t = 20

20 - 7 = 14

Thus, The answer is 14

-TheUnknownScientist 72

5 0

3 years ago

Read 2 more answers

Samuel helps his brother paint walls during his summer vacation he charges $10 for the first room and $5 for each additional roo

Delicious77 [7]

10+5x=50
subtract ten from both sides
5x=40
divide five from both sides
x=8
samuel will need to paint 8 additional rooms

5 0

3 years ago

PLEASE HELP! (WILL GIVE BRAINLIEST FOR RIGHT ANSWER)

dimaraw [331]

Answer:

-41.29* 10^-7

Step-by-step explanation:

You would change the order of the numbers by multiplying -4.44 and 9.3 together and multiply 10^-7 and 10^0. Which would now look like this:

(-4.44* 9.3)* (10^-7* 10^0)

-4.44* 9.3= -41.292 and 10^-7* 10^0= 1* 10^-7. So now the equation would look like this:

-41.292* 10^-7

Then round -41.292 to 2 decimal places which would be -41.29.

-41.29* 10^-7

That's the answer. I hope this helps!

6 0

3 years ago

The owner of a small restaurant bought 75 kilograms of rice. Each week, the restaurant uses 4.5 kilograms of rice.

leva [86]

Answer:

6 weeks 27kilo

12 weeks 54 kilo

Step-by-step explanation:

4.5*12=54

4.5*6=27

3 0

3 years ago

Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro

Vinvika [58]

Answer:

40.1% probability that he will miss at least one of them

Step-by-step explanation:

For each target, there are only two possible outcomes. Either he hits it, or he does not. The probability of hitting a target is independent of other targets. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

0.95 probaiblity of hitting a target

This means that $p = 0.95$

10 targets

This means that $n = 10$

What is the probability that he will miss at least one of them?

Either he hits all the targets, or he misses at least one of them. The sum of the probabilities of these events is decimal 1. So

$P(X = 10) + P(X < 10) = 1$

We want P(X < 10). So

$P(X < 10) = 1 - P(X = 10)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{10,10}.(0.95)^{10}.(0.05)^{0} = 0.5987$

$P(X < 10) = 1 - P(X = 10) = 1 - 0.5987 = 0.401$

40.1% probability that he will miss at least one of them

7 0

3 years ago