Help me please -no wrong answer = no Brainy, right answer = Brainiest

the time t required to cut the grass in a park varies inversely as the number of workers cutting the grass w. If 6 workers can c

mafiozo [28]

24 = 3w
w = 8 workers

3 0

2 years ago

I need help ASAP!!!!!

kupik [55]

Answer:

y = 50

Step-by-step explanation:

Firstly, let's make an equation. We know that when y is 15, x is 3. Or, in other words, y is 5 times what x is. Therefore we can use the equation, y = 5x.

Now, just plug x into the equation.

y = 5(10)

y = 50

5 0

2 years ago

Read 2 more answers

Which of the following expressions is equal to 1 divided by 16?

belka [17]

I wrote it somewhere else where I can’t copy and paste so hope this works!

8 0

2 years ago

Suppose you want to buy one pair of pants and several pairs of socks.

dedylja [7]

Answer:

8

Step-by-step explanation:

50/5.95

3 0

2 years ago

The following prices, in dollars, of 7.5-cubic-foot refrigerators were recorded from a random sample. 314 305 344 283 285 310 3

Mariana [72]

Answer:

$t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109$

The degrees of freedom are given by:

$df=n-1=10-1=9$

And the p value would be given by:

$p_v =P(t_{9}>1.109)=0.148$

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

Step-by-step explanation:

Information given

314 305 344 283 285 310 383 285 300 300

We can calculate the sample mean and deviation with the following formula:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s =\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}$

$\bar X=310.9$ represent the sample mean

$s=31.09$ represent the standard deviation for the sample

$n=10$ sample size

$\mu_o =300$ represent the value to test

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to verify if the true mean is greater than 300, the system of hypothesis would be:

Null hypothesis: $\mu \leq 300$

Alternative hypothesis: $\mu > 300$

The statistic is given by:

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

Replacing the info given we got:

$t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109$

The degrees of freedom are given by:

$df=n-1=10-1=9$

And the p value would be given by:

$p_v =P(t_{9}>1.109)=0.148$

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

6 0

2 years ago