What is AC? Giving brainliest

Yesterday, Gordon had 128 baseball cards. Today, he got d more. Using d, write an expression for the total number cards he has n

raketka [301]

Answer: he has 128 + d

Step-by-step explanation:

well if he had 128 yesterday and today he got d more just add d to 128 and ull see

for example, lets say d = 16

to find out the total we woukd have to add 16 to 128 (just an example the answer is not d=16)

answer= 128+d

3 0

2 years ago

!Can some1 help! !Help FAST!

NemiM [27]

Answer:

4y-9

Step-by-step explanation:

y is the number, 4 times the number = 4y

so 4y-9

4 0

3 years ago

Read 2 more answers

Issac made a mistake in his checkbook. he wrote a check for $8.98 to rent a video game but mistakenly recorded it in his checkbo

Naya [18.7K]

Debit is where you have to pay extra when you use it. Credit doesn't!!!!!

3 0

3 years ago

Concerns about the climate change and CO2 reduction have initiated the commercial production of blends of biodiesel (e.g. from r

natta225 [31]

Answer:

a) 99% of the sample means will fall between 0.933 and 0.941.

b) By the Central Limit Theorem, approximately normal, with mean 0.937 and standard deviation 0.0015.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .