HELP ME FAST PLEASE

Susan decided to start a bracelet business. She spent $17.75 on supplies for her

Snezhnost [94]

Answer:

Given that:

Susan spent on supplies = $17.75

Earning of june = 24 * $3.25

Earning of June = $78

1/4 of earning = 1/4 * 78

1/4 of earning = $19.5

So Susan will set $19.5 aside for supplies of next month and remaining amount she will have is : 78 - 19.5 = $58.5

i hope it will help you!

6 0

3 years ago

FIRST CORRECT GETS BRAINLIEST!!!!!!!! 20 POINTS!!! ASAP!!!!!!!!!!!!!!!!!

den301095 [7]

Answer:

so he can get brainliest ^^

also I agree with the third option

hope this helps :)

5 0

3 years ago

Read 2 more answers

Janine inflated a ball to a radius of 18cm and another ball to a radius of 12cm. how much greater was the volume of air in the l

umka2103 [35]

Answer:

17,190.79 cm ³

Step-by-step explanation:

24429.02 - 7238.23

4 0

3 years ago

A rectangle is placed around a semicircle as shown below. The width of the rectangle is 9 mm. Find the area of the shaded region

kogti [31]

Answer:

Step-by-step explanation:

So since 9 is your radius because this is half a circle you want to use the equation piR^2 which is pi9^2, by solving this equation you end up with 254.34, which is the area of the semi-circle, now you need to find the area of the box but to solve this you would need the length and width and multiply them together to get the area of the box then subtract that by the area of the semi-circle and you get the shaded region

4 0

3 years ago

Tensile-strength tests were carried out on two different grades of wire rod (Fluidized Bed Patenting of Wire Rods, Wire J., June

Shtirlitz [24]

Answer:

$t=\frac{(123.6-107.6)-(10)}{\sqrt{\frac{2^2}{129}+\frac{1.3^2}{129}}}=28.569$

$p_v =P(t_{256}>28.569) \approx 0$

So with the p value obtained and using the significance level given $\alpha=0.1$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm2.

Step-by-step explanation:

The statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2 +10$

Alternative hypothesis: $\mu_1 >\mu_2 +10$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 \leq 10$

Alternative hypothesis: $\mu_1 -\mu_2>10$

Our notation on this case :

$n_1 =129$ represent the sample size for group AISI 1078

$n_2 =129$ represent the sample size for group AISI 1064

$\bar X_1 =123.6$ represent the sample mean for the group AISI 1078

$\bar X_2 =107.6$ represent the sample mean for the group AISI 1064

$s_1=2.0$ represent the sample standard deviation for group 1 AISI 1078

$s_2=1.3$ represent the sample standard deviation for group AISI 1064

And now we can calculate the statistic:

$t=\frac{(123.6-107.6)-(10)}{\sqrt{\frac{2^2}{129}+\frac{1.3^2}{129}}}=28.569$

Now we can calculate the degrees of freedom given by:

$df=129+129-2=256$

And now we can calculate the p value using the altenative hypothesis:

$p_v =P(t_{256}>28.569) \approx 0$

So with the p value obtained and using the significance level given $\alpha=0.1$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm2.

7 0

3 years ago