Maka made $45,670 last year.

a jogger ran 4 miles, decreased her speed by 1 mile per hour, and then ran another 5 miles. If her total jogging time was 2 1/20

LekaFEV [45]

Let the speed of the joker initially be x mi/hr
after reduction, the new speed will be: (x-1) mi/hr
but
time=(distance)/(speed)
hence :
Total time taken will be:
4/x+5/(x+1)=41/20
solving for x we get:
x=21.585 mi/hr
thus the initial speed was 21.585 mi/hr after reduction the new speed was 20.585 mi/hr

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=5%20%5Csqrt%7B15%7D%20" id="TexFormula1" title="5 \sqrt{15} " alt="5 \sqrt{15} " align="absmid

PSYCHO15rus [73]

The answer is 1.71......

5 0

3 years ago

The mean SAT score in mathematics, M, is 600. The standard deviation of these scores is 48. A special preparation course claims

kupik [55]

Answer:

Step-by-step explanation:

The mean SAT score is $\mu=600$ , we are going to call it \mu since it's the "true" mean

The standard deviation (we are going to call it $\sigma$ ) is

$\sigma=48$

Next they draw a random sample of n=70 students, and they got a mean score (denoted by $\bar x$ ) of $\bar x=613$

The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.

- So the Null Hypothesis $H_0:\bar x \geq \mu$

- The alternative would be then the opposite $H_0:\bar x < \mu$

The test statistic for this type of test takes the form

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}$

and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.

With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}\\\\= \frac{| 600-613 |}{48/\sqrt(70}}\\\\= \frac{| 13 |}{48/8.367}\\\\= \frac{| 13 |}{5.737}\\\\=2.266\\$

<h3>since 2.266>1.645 we can reject the null hypothesis.</h3>

6 0

3 years ago

Read 2 more answers

Me need more help plz also add explanation real explanation

Reika [66]

So we want to find the surface area of this pyramid. Finding the surface area of a 3D shape is essentially finding the area of all its sides and adding them together.

What we're going to have to do here is find the area of the base of the pyramid, which is basically a square, and find the area of all four sides of the pyramid, which are triangles.

Formula for area of a square: length × width

Formula for area of a triangle: (base × height) ÷ 2

First we will find the area of the base:

l × w = A

(Write out formula)

8 × 8 =

(Input values)

A = 64 m^2 (square meters or meters squared)

(We have our answer, dont forgot to write your units.)

So we've figured out the area of the base. Now onto the triangles:

(b×h) ÷ 2 = A

(Write out formula)

(8 × 6) ÷ 2 =

(Input values)

48 ÷ 2 =

(Use PEMDAS to solve this correctly. We did whatever was in the parenthesis first, which was multiplication.)

A = 24m^2 (square meters or meters squared)

(We have our answer.)

Now we've found the area of the base of the pyramid and one of the sides of the pyramid, but we're almost done! There's four sides to the pyramid, they're all the same. So since we found the area for one, we can multiply that by 4 to find the area of all sides added together:

24 × 4

(area of a side × total sides)

= 96m^2

All that's left is adding the 96 to the area of the base to get our total surface area of the pyramid:

96 + 64 = 160

The surface area of the pyramid is 160m^2(square meters or meters squared)

(Hope this helps :) )

8 0

3 years ago

Read 2 more answers

X-3y=7 and 2x-6y=12 solving by elimination

Ber [7]

x-3y=7 2x-6y=12

You can multiply -2 in the first equation like this.

x+6y=7

2x-6y=12

Now you can eliminate 6y and -6y so cross them out and do x-2 which is -1 and 7-12 which is -5 -1/-5 would be .2

<h3 />

7 0

3 years ago