How do you solve -4+√(6k+96)=k

Can get some help on 2,4,22, 42 and 50 please

Fittoniya [83]

Answer:

2. zeros

4. difference

22. 79,158.5

42. 0.0099

50. 4.37

Step-by-step explanation:

2. Only adding 0s after a decimal point does not change the value.

4. Since you are subtracting then you estimate difference.

22. Add each number using place value.

75285

2

3971

0.5

__________

79258.5

42. 0.01 is really 0.0100. If you subtract 0.0001 then it is like subtracting 100-1 = 99 but in the decimal place. So it becomes 0.0099.

50. Subtract by lining up decimal places. Then move place values to subtract.

9.00

- 4.63

----------------

4.37

6 0

3 years ago

A recent study reported that high school students spend an average of 94 minutes per day texting. Jenna claims that the average

larisa [96]

Answer:

We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.

Step-by-step explanation:

Jenna claims that the average time of texting at her larger high school is greater than 94 minutes per day.

From here we can see that we have to perform a hypothesis test about a sample mean. The null and alternate hypothesis will be:

Null Hypothesis: $\mu \leq 94$

Alternate Hypothesis: $\mu > 94$

Jenna collected data from a sample of 32 students. So, sample size will be:

Sample Size = n = 32

Sample Mean = x = 96.5

Sample Standard Deviation = s = 6.3

We have to perform a hypothesis test, to test Jenna's claim. Since, the value of Population Standard Deviation is unknown and the value of Sample Standard Deviation is known, we will use One Sample t-test in this case.

The formula to calculate the test statistic is:

$t=\frac{x-\mu}{\frac{s}{\sqrt{n}}}$

Using the values, we get:

$t=\frac{96.5-94}{\frac{6.3}{\sqrt{32} } }=2.245$

The degrees of freedom will be:

df = n - 1 = 32 - 1 = 31

We have to convert the t-score 2.245 with 31 degrees of freedom to its equivalent p-value. From t-table this value comes out to be:

p-value = 0.0160

The significance level is:

$\alpha =0.05$

Since, the p-value is lesser than the level of significance, we reject the Null Hypothesis.

Conclusion:

We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.

3 0

3 years ago

One shape is approaching a poor from the east traveling west at 15 mph and its presently 3 miles east east of the port I second

valentinak56 [21]

Answer:

DL/dt = 17 mph

Step-by-step explanation:

The ships and the port shape a right triangle

Ship going west ( x-direction) is traveling at 15 mph

Ship going north (y-direction) is traveling at 10 mph

The distance L between ships is:

L² = x² + y²

Tacking derivatives on both sides of the equation with respect to time

we get

2*L*DL/dt = 2*x*Dx/dt + 2*y*Dy/dt (1)

In that equation we know:

Dx/dt = 15 mph

Dy/dt = 10 mph

At the moment ship traveling from the east is at 3 miles from the port

then x = 3 m and the other ship is at 4 miles north

then by Pythagoras theorem

L = √ 3² + 4² L = 5

By substitution in equation 1

2*5*DL/dt = 2*3*15 + 2*4*10

10* DL/dt = 90 + 80

DL/dt = 170 / 10

DL/dt = 17 mph

4 0

3 years ago

Width of a house cm dm m or km

ziro4ka [17]

Well it depends on what the units say in the problem you are looking at, if it is not the use 'square units' or 'units'

3 0

3 years ago

Firstright answer will get brainlist

avanturin [10]

P:(1/2)=50
p=50*(1/2)=25

5 0

3 years ago