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3 years ago

Teagan was playing Trouble and recorded the number

Mathematics

1 answer:

almond37 [142]3 years ago

4 0

Answer:

20%

Step-by-step explanation:

Think about it. 100 divided by 25 is four so just multiply 5 times 4 and you got 20. giving you the probability of 20%. Your Welcome! And I'm not a genius it's just simple math.

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A teacher gives a test in which the average of the test scores is 68 and the standard deviation is 14. Realizing that the test w

Alika [10]

Answer:

Mean = 78 ;

Standard deviation = 14

Step-by-step explanation:

Given that:

Mean score = 68

Standard deviation = 14

When a constant value is added to the individual value of a dataset, the new mean value will become the sum of the initial mean value and the constant term added to the values.

Constant Added score = 10

The mean of scaled test score will be:

Mean score + constant added score

68 + 10 = 78

The standard deviation remains unchanged ; Hence, standard deviation = 14

6 0

3 years ago

PLEASE HELP WILL GIVE BRAINLIEST ANSWER!!

andreyandreev [35.5K]

Speed = Distance / Time

Distance = 5t^4 - 10t^2+6

Time = t+2

Now:

5t^4 - 10t^2 + 6 | t+2
-5t^4 -10t^3 5t^3 - 10t^2 + 10t - 20
\\\\\ -10t^3-10t^2+6
+10t^3+20t^2
\\\\\\\ 10t^2+6
-10t^2-20t
\\\\\ -20t+6
+20t+40
\\\\\ 46

<span>C. 5t^3-10t^2+10t-20+(46)/(t+2))</span>

3 0

3 years ago

If i walk 4 miles per hour how long will it take to walk a mile

fiasKO [112]

Hi there! The answer is 15 minutes.

Walking 4 miles per hour means that it takes you 60 minutes to walk 4 miles. To find the time it takes you to walk 1 mile, we must divide both sides by 4.

4 miles equals 60 minutes.
1 mile equals 60 / 4 = 15 minutes.

The answer is 15 minutes.

4 0

4 years ago

Given: PS=RT, PQ=ST<br> Prove: QS=RS

ivanzaharov [21]

Answer:

I) Eq(1) reason: sum of segments of a straight line

II) Eq(2) reason: Given PQ = ST & PS = RT

III) Eq(3) reason: sum of segments of a straight line

IV) Eq(4) reason: Same value on right hand sides of eq(2) and eq(3) demands that we must equate their respective left hand sides

V) Eq(5) reason: Usage of collection of like terms and subtraction provided this equation.

Step-by-step explanation:

We are given that;

PS = RT and that PQ = ST

Now, we want to prove that QS = RS.

From the diagram, we can see that from concept of sum of segments of a straight line we can deduce that;

PQ + QS = PS - - - - (eq 1)

Now, from earlier we saw that PQ = ST & PS = RT

Thus putting ST for PQ & PS for RT in eq 1,we have;

ST + QS = RT - - - - (eq 2)

Again, from the line diagram, we can see that from concept of sum of segments of a straight line we can deduce that;

RS + ST = RT - - - - -(eq 3)

From eq(2) & eq(3) we can see that both left hand sides is equal to RT.

Thus, we can equate both left hand sides with each other to give;

ST + QS = RS + ST - - - (eq 4)

Subtracting ST from both sides gives;

ST - ST + QS = RS + ST - ST

This gives;

QS = RS - - - - (eq 5)

Thus;

QS = RS

Proved

5 0

3 years ago

A lathe is set to cut bars of steel into lengths of 6 cm. The lathe is considered to be in perfect adjustment if the average len

Gnom [1K]

Answer:

$t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723$

E. -0.723

$df=n-1=93-1=92$

$p_v =2*P(t_{(92)}$

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.

Step-by-step explanation:

Information provided

$\bar X=5.97$ represent the sample mean for the length

$s=0.4$ represent the sample standard deviation

$n=93$ sample size

$\mu_o =6$ represent the value that we want to test

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value for the test

System of hypothesis

We need to conduct a hypothesis in order to check if the lathe is in perfect adjustment (6cm), then the system of hypothesis would be:

Null hypothesis: $\mu = 6$

Alternative hypothesis: $\mu \neq 6$

since we don't know the population deviation the statistic is:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing in formula (1) we got:

$t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723$

E. -0.723

P value

The degrees of freedom are given by:

$df=n-1=93-1=92$

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

$p_v =2*P(t_{(92)}$

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.

5 0

3 years ago