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3 years ago

If 5x+6-19=5 what is x

Mathematics

2 answers:

Troyanec [42]3 years ago

5 0

Answer:

<em>x </em>equals 3.6.

Step-by-step explanation:

A way to solve this equation is to preform the inverse operations on both sides.

1. Add 19 to -19 and 5: 5<em>x </em>+ 6 - 19 + 19 = 5 + 19 ---> 5x + 6 = 24

2. Subtract 6 from 6 and 5: 5x + 6 - 6 = 24 - 6 ---> 5x = 18

3. Divide 5x and 18 by 5: 5x / 5 = 18 / 5 ---> x = 3.6

4. 3.6 is the answer.

To prove it, substitute x with 3.6 in the equation: 5 · 3.6 + 6 - 19 = 5

I hope this made sense and helped you a lot! :)

Effectus [21]3 years ago

5 0

Answer:
x equals to 3.6

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What is 9.562359123571283578 rounded to the nearest hundredth?

Katyanochek1 [597]

9.56 because you don't round up when two is the next number. (The hundredths place is the second number after the decimal, where the 6 is in the original number.)

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3 years ago

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Could these triangles be congruent?

Mama L [17]

Answer:

yes, if AB ≅ DE

Step-by-step explanation:

Triangles are said to be congruent if they have the same sides and the same angles.

The following measures are used to determine if triangles are congruent:

1) Angle-side-angle: If two angles and a side of a triangle is equal to two angles and corresponding side of another triangle, then they are congruent.

2) Side-side-side: If all three sides of a triangle is equal to three sides of another triangle, then the two triangles are congruent.

3) Side angle side: If two sides and an included angle of a triangle is equal to the two sides and corresponding angle of another triangle, then they are congruent.

4) Hypotenuse - leg: If the hypotenuse and one leg of a triangle is equal to the hypotenuse and leg of another triangle then they are congruent.

From the triangles DEF and ABC, already, they already have one equal triangle that is ∠F = ∠B and an equal side i.e DF = AC.

To satisfy congruence, two sides and an angle have to be equal, therefore if AB = DE then the two triangles would be congruent

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3 years ago

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Toilet Training You are a great friend and are taking your friend's child to the playground to play with two of his friends: Cha

Brilliant_brown [7]

Answer:

Step-by-step explanation:

Hello!

According to a study regarding the average age of female and male kids to complete toilet training is:

Females:

Average age 32.5 months

The standard deviation of 6.7 months

n= 126

Males:

Average age 35.0 months

The standard deviation of 10.1 months

n= 141

The parameter of study is the difference between the age of females are successfully toilet trained and the average age that males are successfully toilet trained. μf - μm (f= female and m= male)

Assuming that both variables have a normal distribution to choose whether you'll use an unpooled or pooled-t to calculate the confidence interval you have to conduct an F-test for variance homogeneity.

If the variances are equal, then you can usee the pooled-t, but if the variances are different, you have to uses Wlche's approach:

H₀: δ²f = δ²m

H₁: δ²f ≠ δ²m

Since both items b. and c. ask for a 95% CI I'll use the complementary significance level for this test:

α: 0.05

$F= \frac{S^2_f}{S^2_m} * \frac{xSigma^2_f}{Sigma^2_m} ~~~F_{(n_f-1); (n_m-1)}$

$F= \frac{(6.7^2)}{(10.1)^2} *1= 0.44$

Critical values:

$F_{125;140;0.025}= 0.71\\F_{125;140;0.975}= 1.41$

The calculated F value is less than the lower critical value, 0.77, so the decision is to reject the null hypothesis. In other words, there is no significant evidence to conclude the population variances of the age kids are toilet trained to be equal. You should use Welch's approach to construct the Confidence Intervals.

$Df_w= \frac{(\frac{S^2_f}{n_f} + \frac{S^2_m}{n_m} )^2}{\frac{(\frac{S^2_f}{n_f} )^2}{n_f-1} +\frac{(\frac{S^2_m}{n_m} )^2}{n_m-1} }$

$Df_w= \frac{(\frac{6.7^2}{126} + \frac{10.1^2_m}{141} )^2}{\frac{(\frac{126^2}{126} )^2}{126-1} +\frac{(\frac{10.1^2}{141} )^2}{141-1} } = 254.32$

The given interval is:

[0.3627; 4.6073]

Using Welch's approach, the formula for the CI is:

(X[bar]f- X[bar]m) ± $t_{Df_w;1-\alpha /2}$ * $\sqrt{\frac{S^2_f}{n_f} +\frac{S^2_m}{n_m} }$

(X[bar]m- X[bar]f) ± $t_{Df_w;1-\alpha /2}$ * $\sqrt{\frac{S^2_f}{n_f} +\frac{S^2_m}{n_m} }$

As you can see either way you calculate the interval, it is centered in the difference between the two sample means, so you can clear the value of that difference by:

(Upper bond - Lower bond)/2= (4.6073-0.3627)/2= 2.1223

The average age for females is 32.5 months and for males, it is 35 months.

Since the difference between the sample means is positive, we can say that the boys were considered "group 1" and the girls were considered "group 2"

You have a95% confidence that the parameter of interest is included in the given confidence interval.

None of the statements is correct, the interval gives you information about the difference between the average age the kids are toilet trained, that is between the expected ages for the entire population of male and female babies.

This represents a guideline but is not necessarily true to all individuals of the population since some male babies can be toilet trained before that is expected as some female babies can be toiled trained after the average value.

I hope it helps!

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3 years ago

**TIMED QUIZ PLEASE HELP** Which expression is equivalent to (x+1)^2-9 over x+4?

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Answer:

Step-by-step explanation:

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3 years ago

There is a single sequence of integers $a_2$, $a_3$, $a_4$, $a_5$, $a_6$, $a_7$ such that \[\frac{5}{7} = \frac{a_2}{2!} + \frac

Nataliya [291]

You have a single sequence of integers $a_2,\ a_3,\ a_4,\ a_5,\ a_6,\ a_7$ such that

$\dfrac{a_2}{2!} + \dfrac{a_3}{3!} + \dfrac{a_4}{4!} + \dfrac{a_5}{5!} + \dfrac{a_6}{6!} + \dfrac{a_7}{7!}=\dfrac{5}{7},$

where $0 \le a_i < i$ for $i = 2, 3, \dots, 7.$

1. Multiply by 7! to get

$\dfrac{7!a_2}{2!} + \dfrac{7!a_3}{3!} + \dfrac{7!a_4}{4!} + \dfrac{7!a_5}{5!} + \dfrac{7!a_6}{6!} + \dfrac{7!a_7}{7!}=\dfrac{7!\cdot 5}{7},\\ \\7\cdot 6\cdot 5\cdot 4\cdot 3\cdoa a_2+7\cdot 6\cdot 5\cdot 4\cdot a_3+7\cdot 6\cdot 5\cdot a_4+7\cdot 6\cdot a_5+7\cdot a_6+a_7=6!\cdot 5,\\ \\7(6\cdot 5\cdot 4\cdot 3\cdoa a_2+6\cdot 5\cdot 4\cdot a_3+6\cdot 5\cdot a_4+6\cdot a_5+a_6)+a_7=3600.$

By Wilson's theorem,

$a_7+7\cdot (6\cdot 5\cdot 4\cdot 3\cdoa a_2+6\cdot 5\cdot 4\cdot a_3+6\cdot 5\cdot a_4+6\cdot a_5+a_6)\equiv 2(\mod 7)\Rightarrow a_7=2.$

2. Then write $a_7$ to the left and divide through by 7 to obtain

$6\cdot 5\cdot 4\cdot 3\cdoa a_2+6\cdot 5\cdot 4\cdot a_3+6\cdot 5\cdot a_4+6\cdot a_5+a_6=\dfrac{3600-2}{7}=514.$

Repeat this procedure by $\mod 6:$

$a_6+6(5\cdot 4\cdot 3\cdoa a_2+ 5\cdot 4\cdot a_3+5\cdot a_4+a_5)\equiv 4(\mod 6)\Rightarrow a_6=4.$

And so on:

$5\cdot 4\cdot 3\cdoa a_2+ 5\cdot 4\cdot a_3+5\cdot a_4+a_5=\dfrac{514-4}{6}=85,\\ \\a_5+5(4\cdot 3\cdoa a_2+ 4\cdot a_3+a_4)\equiv 0(\mod 5)\Rightarrow a_5=0,\\ \\4\cdot 3\cdoa a_2+ 4\cdot a_3+a_4=\dfrac{85-0}{5}=17,\\ \\a_4+4(3\cdoa a_2+ a_3)\equiv 1(\mod 4)\Rightarrow a_4=1,\\ \\3\cdoa a_2+ a_3=\dfrac{17-1}{4}=4,\\ \\a_3+3\cdot a_2\equiv 1(\mod 3)\Rightarrow a_3=1,\\ \\a_2=\dfrac{4-1}{3}=1.$

Answer: $a_2=1,\ a_3=1,\ a_4=1,\ a_5=0,\ a_6=4,\ a_7=2.$

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3 years ago