Effie ate 4 peaches. this represents 25% of the bag of peaches. how many peaches were originally in the bag

I'm thinking of two numbers, 12 and another number. 12 and my other number have a GCF of 6 and a LCM of 36. What's the other num

mihalych1998 [28]

Answer:

As GCF is

6

and LCM is

36

, and one number is

12

other number is

6

×

36

12

=

6

×

3

36

1

12

=

18

Step-by-step explanation:

5 0

3 years ago

One airplane located 200km north and 50km east of an airport. A second plane at the same altitude is located 30km north and 100k

grigory [225]

Answer: 350

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

A road perpendicular to a highway leads to a farmhouse located d miles away. An automobile traveling on this highway passes thro

pshichka [43]

Answer:

$\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}$

Step-by-step explanation:

A road is perpendicular to a highway leading to a farmhouse d miles away.

An automobile passes through the point of intersection with a constant speed $\frac{dx}{dt}$ = r mph

Let x be the distance of automobile from the point of intersection and distance between the automobile and farmhouse is 'h' miles.

Then by Pythagoras theorem,

h² = d² + x²

By taking derivative on both the sides of the equation,

$(2h)\frac{dh}{dt}=(2x)\frac{dx}{dt}$

$(h)\frac{dh}{dt}=(x)\frac{dx}{dt}$

$(h)\frac{dh}{dt}=rx$

$\frac{dh}{dt}=\frac{rx}{h}$

When automobile is 30 miles past the intersection,

For x = 30

$\frac{dh}{dt}=\frac{30r}{h}$

Since $h=\sqrt{d^{2}+(30)^{2}}$

Therefore,

$\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+(30)^{2}}}$

$\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}$

3 0

3 years ago

Consider the following hypothesis test:

Gnom [1K]

Answer:

a. z=3.09. Yes, it can be concluded that the population mean is greater than 50.

b. z=1.24. No, it can not be concluded that the population mean is greater than 50.

c. z=2.22. Yes, it can be concluded that the population mean is greater than 50.

Step-by-step explanation:

We have a hypothesis test for the mean, with the hypothesis:

$H_0: \mu\leq50\\\\H_a:\mu> 50$

The sample size is n=55 and the population standard deviation is 6.

The significance level is 0.05.

We can calculate the standard error as:

$\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{6}{\sqrt{55}}=0.809$

For a significance level of 0.05, the critical value for z is zc=1.644. If the test statistic is bigger than 1.644, the null hypothesis is rejected.

a. If the sample mean is M=52.5, the test statistic is:

$z=\dfrac{M-\mu}{\sigma_M}=\dfrac{52.5-50}{0.809}=\dfrac{2.5}{0.809}=3.09$

The null hypothesis is rejected, as z>zc and falls in the rejection region.

b. If the sample mean is M=51, the test statistic is:

$z=\dfrac{M-\mu}{\sigma_M}=\dfrac{51-50}{0.809}=\dfrac{1}{0.809}=1.24$

The null hypothesis failed to be rejected, as z<zc and falls in the acceptance region.

c. If the sample mean is M=51.8, the test statistic is:

$z=\dfrac{M-\mu}{\sigma_M}=\dfrac{51.8-50}{0.809}=\dfrac{1.8}{0.809}=2.22$

The null hypothesis is rejected, as z>zc and falls in the rejection region.

8 0

3 years ago

SOMEONE HELP ME PLEASE QUESTION I'll give you the crown please someone help An equation using at least seven numbers and / or sy

stiv31 [10]

Answer:

7+9-4/2-7+21=28

Step-by-step explanation:

8 0

3 years ago