Solve this x/5 = 2 1/2

5×100+9×1+41/100 as a decimal

liubo4ka [24]

Answer:

Step-by-step explanation:

Parentheses Exponents Multiplication Division Addition Subtraction

5X100=500 and 9X1=9 and 41/100=0.41 Then put them back into the equation

5X100+9X1+41/100=

500 + 9 + 0.41 = Now add. 509.41

3 0

3 years ago

Please help me on number 4

Mariulka [41]

Angle 5 = 117° because it is vertical to angle 8
Angle 7 = 63° because it is symmetrical to angle 5
Angle 6 = 63° because it is vertical to angle 7.
Angle 1 = 117° because it is corresponding to angle 5.
Angle 2 = 63° because it is corresponding to angle 6.
Angle 3= 117° because it is corresponding to angle 8.
Angle 4 = 63° because it is corresponding to angle 7.

3 0

3 years ago

Help me please please please

avanturin [10]

Answer:

3.33lbs. repeating?

Step-by-step explanation:

I think it would just be 10/3 and be 3.33333333 and so on. (im just taking a guess)

3 0

3 years ago

Tamika has a spinner with 5 equal sections- red, blue, yellow, and purple. she plans to pin it 150 times predict how many times

Alexxandr [17]

Number times landing on green) = 150 * 1/5 = 30

Number times landing on yellow = 150 * 1/5 = 30

Answer 60 times

6 0

3 years ago

A student at a four-year college claims that mean enrollment at four-year colleges is higher than at two-year colleges in the Un

WINSTONCH [101]

Answer:

Part 1: The statistic

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

Since is a right tailed test the p value would be:

$p_v =P(t_{68}>4.31)=0.000022 \approx 0.00002$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

Step-by-step explanation:

Data given

$\bar X_{1}=5135$ represent the mean for four year college

$\bar X_{2}=4436$ represent the mean for two year college

$s_{1}=783$ represent the sample standard deviation for four year college

$s_{2}=553$ represent the sample standard deviation two year college

$n_{1}=35$ sample size for the group four year college

$n_{2}=35$ sample size for the group two year college

$\alpha=0.01$ Significance level provided

t would represent the statistic (variable of interest)

System of hypothesis

We need to conduct a hypothesis in order to check if the mean enrollment at four-year colleges is higher than at two-year colleges in the United States , the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We can assume that the normal distribution is assumed since we have a large sample size for each case n>30. So then the sample mean can be assumed as normally distributed.

Part 1: The statistic

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

Since is a right tailed test the p value would be:

$p_v =P(t_{68}>4.31)=0.000022$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

6 0

3 years ago