Does anyone know the answer to this question?

210 divided by 30 please give step my step explanation

Lena [83]

<h2>7</h2><h3>Consider how many times 30 goes into 210(exactly 7 times) and that will be your answer.</h3>

5 0

3 years ago

Please help it’s about to be due

lana66690 [7]

Answer:

D.Z

u see how they have line in between as well, shows that they are going by 5 so 2.0 and then 2.5

and if it was asking for -2.5 which is negative two point five, you would choose A.U

8 0

3 years ago

Find the constant of proportionality for the table of values

elena-14-01-66 [18.8K]

The proportionality for the table values would be an increasing value of 1.5y for every X.

3 0

4 years ago

The friendly sausage factory (fsf) can produce hot dogs at a rate of 5,000 per day. fsf supplies hot dogs to local restaurants a

juin [17]

Answer:

a. 21 327 hot dogs/run

b. 70 runs/yr

c. 4 da/run

Step-by-step explanation:

Data:

Production rate (p) = 5000/da

Usage rate (u) = 260/da

Setup cost (S) = $66

Annual carrying cost (H) = $0.45/hot dog

Production days (d) = 294 da

Calculations:

a. Optimal run size

(i) Annual demand (D) = pd = (5000 hot dogs/1 day) × (294 days/1 yr)

= 1 470 000 hot dogs/yr

(ii) Economic run size

$Q_{0}= \sqrt{\frac{2DS }{ h}\times\frac{ p}{p-u }}$

$= \sqrt{\frac{2\times1470000\times66 }{ 0.45}\times\frac{ 5000}{5000-260 }}$

$= \sqrt{431200000\times\frac{ 5000}{4740 }}$

$= \sqrt{454852321}$

= 21 327 hot dogs/run

b. Number of runs per year

Runs = D/Q₀ = (1 470 000 hot dogs/1yr) × (1 run/21 327 hotdogs)

= 70 runs/yr

c. Length of a run

Length = Q₀/p = (21 327 hot dogs/1 run) × (1 da/5000 hot dogs)

= 4 da/run

8 0

3 years ago

A car company claims that its cars achieve an average gas mileage of at least 26 miles per gallon. A random sample of eight cars

motikmotik

Answer:

$t=\frac{25.5-26}{\frac{1}{\sqrt{8}}}=-1.414$

The degrees of freedom are given by:

$df=n-1=8-1=7$

The p value for this case is given by:

$p_v =P(t_{(7)}$

Since the p value is higher than the significance level of 0.06 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 25.5 and then the claim makes sense

Step-by-step explanation:

Information given

$\bar X=25.5$ represent the sample mean

$s=1$ represent the sample standard deviation

$n=8$ sample size

$\mu_o =26$ represent the value to verify

$\alpha=0.06$ represent the significance level

t would represent the statistic (variable of interest)

$p_v$ represent the p value

Hypothesis to est

We want to test if the true mean is at least 26 mpg, the system of hypothesis would be:

Null hypothesis: $\mu \geq 25.5$

Alternative hypothesis: $\mu < 25.5$

The statistic for this case is given by;

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

Replacing the info given we got:

$t=\frac{25.5-26}{\frac{1}{\sqrt{8}}}=-1.414$

The degrees of freedom are given by:

$df=n-1=8-1=7$

The p value for this case is given by:

$p_v =P(t_{(7)}$

Since the p value is higher than the significance level of 0.06 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 25.5 and then the claim makes sense

3 0

4 years ago