Which expression shows that 11 + 7 = 5 + 13?

Can someone help I been stuck on the question of 10min

shusha [124]

Answer:

y e s ∩ω∩

Step-by-step explanation:

(I think it is correct)

5 0

2 years ago

Read 2 more answers

What is y=-10x's constant of proportionality and how would describe a graph's constant rate of change.

Fantom [35]

Okay, so here, we know that -10 is the slope, therefore, it is also the constant of proportionality.

'x' is the unknown value, that tells us to multiply by -10.

So, on a graph, we would consider that per unit, it would decrease by -10, since its a negative slope.

Hope I helped, if you have further questions or concerns, feel free to PM me. Thanks! :D

5 0

3 years ago

The table shows the profit and loss of the after -school snack stand write each profit as a positive number and each loss a nega

kiruha [24]

Answer:

Profit

7.50

4.50

Loss

-3.50

-6.00

Step-by-step explanation:

Here, we want to write each profit as a positive number and each loss as a negative number

thus, we have that;

Profit;

7.50

4.50

Loss;

-3.50

-6.00

7 0

2 years ago

Gabriella swims 2 laps per minute. What is the associated rate? ___ minute per lap

Leokris [45]

Answer:

The answer is D.

Step-by-step explanation:

It is given that Gabriella swims 2 laps for 1 minute. So in order to find how many minutes did he swim for each lap, you have to divide it by 2 :

2 laps = 1 minute

2 laps ÷ 2 = 1 minute ÷ 2

1 lap = 1/2 minute

= 30 seconds

Test :

Gabriella swim 1/2 min per lap. So if he swim 2 laps, you have to multiply it by 2 :

1/2 × 2 = 1 minute

7 0

2 years ago

Read 2 more answers

A manufacturer uses production method to produce steel rods. A random sample of 17 steel rods resulted in lengths with a standar

Arisa [49]

Answer:

$\chi^2 =\frac{17-1}{12.15} 22.09 =15.87$

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates $\alpha/2 =0.05$ of the area on each tail and we got:

$\chi^2_{crit}= 7.962$

$\chi^2_{crit}=26.296$

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is NOT significantly different from 3.5 cm

Confidence level method

We can find the p value like this

$p_v =2*P(\chi^2 >15.87)=0.92$

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is NOT significantly different from 3.5 cm

Step-by-step explanation:

Data provided

$n=17$ represent the sample selected

$\alpha=0.1$ represent the significance

$s^2 =4.7^2 =22.09$ represent the sample variance

$\sigma^2_0 =3.5^2 =12.15$ represent the value to check

Null and alternative hypothesis

We want to verify if the new production method has lengths with a standard deviation different from 3.5 cm, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 = 12.15$

Alternative hypothesis: $\sigma^2 \neq 12.15$

The statistic for this case is given by:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

The degrees of freedom are:

$df =n-1= 17-1=16$

$\chi^2 =\frac{17-1}{12.15} 22.09 =15.87$

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates $\alpha/2 =0.05$ of the area on each tail and we got:

$\chi^2_{crit}= 7.962$

$\chi^2_{crit}=26.296$

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is significantly different from 3.5 cm

Confidence level method

We can find the p value like this

$p_v =2*P(\chi^2 >15.87)=0.92$

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is significantly different from 3.5 cm

7 0

3 years ago