What percent is equivalent to 3/10

grigory [225]

0.86/5-0.3*0.5= 2 answers

1. can be 11/500

2. it can be 0.022 but im positive it has to be 11/500

5 0

3 years ago

Ned,Tony,Matt and juan are palyung basketball. Ned scores 2p+3 points more than Ned, Matt scored teice as many points as Tony, a

horsena [70]

<span>Ned scored 2p+3. Tony scored 3 points more than Ned, so Tony scored 3+2p+3, or 2p+6. Matt scored twice as many points as Tony, so Matt scored 2(2p+6), or 4p+12. Juan scored 8 fewer points than Ned, so Juan scored 2p+3-8, or 2p-5. To simplify, we add these four values. 2p + 3 2p + 6 4p + 12 2p - 5 This equals 10p + 16.</span>

7 0

3 years ago

Help show work pls for brisnly

Deffense [45]

Answer:

6

Step-by-step explanation:

Ok, so all we have to do is multiply 8 * 3/4, because each child needs 3/4 of the sandwich, and there are 8 of them. The easiest (My opinion) way to do this is by turning 8 into a fraction. Turning whole numbers into fractions is really simple, just slap a /1 under them to make them a fraction. And so, our fraction is 8/1. Now, all we have to do is multiply the top part of the fraction, and then the bottom part of the fraction. So, 8 * 3 = 24, and 1 * 4 = 4. So, just put 24/4 as an answer. But, wait! We have to simplify it! 24 divided by 4 is 6, so 6 is our final answer.

8 0

3 years ago

A variety of stores offer loyalty programs. Participating shoppers swipe a bar-coded tag at the register when checking out and r

dsp73

Answer:

$t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236$

Now we can find the degrees of freedom:

$df=n-1=80-1=79$

Now we can calculate the p value with the alternative hypothesis using the following probability:

$p_v =P(t_{(79)}>2.236)=0.0141$

Using a significance level of 0.01 or 1% we have enough evidence to FAIL to reject the null hypothesis and we can conclude that shoppers participating in the loyalty program NOT spent more on average than typical shopper at this significance level assumed

Step-by-step explanation:

Information given

$\bar X=130$ represent the sample mean

$s=40$ represent the sample standard deviation

$n=80$ sample size

$\mu_o =120$ represent the value to verify

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to verify if shoppers participating in the loyalty program spent more on average than typical shoppers (120) , the system of hypothesis would be:

Null hypothesis: $\mu \leq 120$

Alternative hypothesis: $\mu > 120$

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:

$t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236$

Now we can find the degrees of freedom:

$df=n-1=80-1=79$

Now we can calculate the p value with the alternative hypothesis using the following probability:

$p_v =P(t_{(79)}>2.236)=0.0141$

Using a significance level of 0.01 or 1% we have enough evidence to FAIL to reject the null hypothesis and we can conclude that shoppers participating in the loyalty program NOT spent more on average than typical shopper at this significance level assumed

3 0

4 years ago

Solve y=3bx-7x for x

irina [24]

Y = 3bx - 7x
y = x(3b - 7)

Divide each side by 3b - 7 (assume that it is not zero).
$\frac{y}{3b-7} =x$

Answer:
$x= \frac{y}{3b-7}$

6 0

3 years ago