If the divisor is 30, what is the least 3 digit dividend that will give you a remainder of 8?

atroni [7]

Answer:

I believe the question says "5j + 16 - 12(-9j - 4) - 12j" I hope I'm correct because I have almost no idea what you typed.

If you did type "5j + 16 - 12(-9j - 4) - 12j" then your answer would be 133j + 52 I believe. You never put how you would like your problem to be solved, so I simplified.

Step-by-step explanation:

Hope this helps.

5 0

3 years ago

The weights of soy patties sold by Veggie Burgers Delight are normally distributed. A random sample of 15 patties yields a mean

Elden [556K]

Answer:

$t=\frac{3.8-4}{\frac{0.5}{\sqrt{15}}}=-1.549$

Step-by-step explanation:

Data given and notation

$\bar X=3.8$ represent the sample mean

$s=0.5$ represent the sample standard deviation for the sample

$n=15$ sample size

$\mu_o =4$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean weight is less than 4 ounces, the system of hypothesis would be:

Null hypothesis: $\mu \geq 68$

Alternative hypothesis: $\mu < 4$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{3.8-4}{\frac{0.5}{\sqrt{15}}}=-1.549$

3 0

3 years ago

What is 2 7/4 in simplest form

Leona [35]

2 7/4=3.75

THE ANSWER IS 3.75

4 0

4 years ago

Read 2 more answers

Find the total amount you will pay on a bill of $43.50 and tip rate of 12%.

Mars2501 [29]

About $48.72

Explanation:

8 0

3 years ago

A local car dealer claims that 25% of all cars in San Francisco are blue. You take a random sample of 600 cars in San Francisco

sammy [17]

Answer:

No, we can't reject the dealer's claim with a significance level of 0.05.

Step-by-step explanation:

We are given that a local car dealer claims that 25% of all cars in San Francisco are blue.

You take a random sample of 600 cars in San Francisco and find that 141 are blue.

Let p = proportion of all cars in San Francisco who are blue

SO, Null Hypothesis, $H_0$ : p = 25% {means that 25% of all cars in San Francisco are blue}

Alternate Hypothesis, $H_A$ : p $\neq$ 25% {means that % of all cars in San Francisco who are blue is different from 25%}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of 600 cars in San Francisco who are blue = $\frac{141}{600}$ = 0.235

n = sample of cars = 600

So, test statistics = $\frac{0.235-0.25}{{\sqrt{\frac{0.235(1-0.235)}{600} } } } }$

= -0.866

Now at 0.05 significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics lies within the range of critical values of z so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that 25% of all cars in San Francisco are blue which means the dealer's claim was correct.

4 0

3 years ago