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2 years ago

Please help me please please please

Mathematics

2 answers:

marishachu [46]2 years ago

8 0

Answer:

√121 = 11

√81 = 9

√4 = 2

√100 = 10

√144 = 12

√16 = 4

√1 = 1

√49 = 7

√25 = 5

√9 = 3

Please mark me as brainliest ❤️

Andrei [34K]2 years ago

6 0

Answer:wanna hang

Step-by-step explanation:

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4(x-6)<-2x+6 what is the solution to the inequality _

Cloud [144]

Simplify both sides if needed. The left-hand side needs simplification.

4(x - 6) $\leq$ -2x + 6
4x - 24 $\leq$ -2x + 6

All is left to do is add and subtract to get the x variable all alone.

4x - 24 $\leq$ -2x + 6
6x - 24 $\leq$ 6 <-- Add 2x to both sides
6x $\leq$ 30 <-- Add 24 to both sides
x $\leq$ 5 <-- Divide both sides by 6

In order to be in the solution set, x has to be less than or equal to 5.

In interval notation: [5, -∞)

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3 years ago

Given the quadratic function y=5x^2−2x identify a, b and c.

saw5 [17]

Work shown above! Answer is
a = 5 b = -2 c = 0

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3 years ago

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Assume that you have a sample of n 1 equals 6, with the sample mean Upper X overbar 1 equals 50, and a sample standard deviati

tigry1 [53]

Answer:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

$df=6+5-2=9$

$p_v =P(t_{9}>2.656) =0.0131$

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

$n_1 =6$ represent the sample size for group 1

$n_2 =5$ represent the sample size for group 2

$\bar X_1 =50$ represent the sample mean for the group 1

$\bar X_2 =38$ represent the sample mean for the group 2

$s_1=7$ represent the sample standard deviation for group 1

$s_2=8$ represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 > \mu_2$

We are assuming that the population variances for each group are the same

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

The statistic for this case is given by:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

The pooled variance is:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

We can find the pooled variance:

$S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67$

And the pooled deviation is:

$S_p=7.46$

The statistic is given by:

$t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656$

The degrees of freedom are given by:

$df=6+5-2=9$

The p value is given by:

$p_v =P(t_{9}>2.656) =0.0131$

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3 years ago

What is the equation of a circle with a center (-2, 3) and a radius r = 5?

qwelly [4]

The formula for the equation of a circle is:
(x - h)^2 + (y - k)^2 = r^2

(h, k) is the center.

So the equation would be:
(x + 2)^2 + (y - 3)^2 = 5^2
or
(x + 2)^2 + (y - 3)^2 = 25

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3 years ago

Pl help it’s for a grade

frosja888 [35]

Answer:

A,d,e not sur tho

Step-by-step explanation:

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2 years ago

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