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mafiozo [28]
2 years ago
14

Solve pls brainliest tell me where the dot go!

Mathematics
1 answer:
maxonik [38]2 years ago
8 0

Answer:

The first dot is on -9 and the second dot is on -7.

(a) -9 is to the left of -7.

(b) -9<-7

Step-by-step explanation:

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Insert parentheses in expression 2 so that it has a value of 4. Then show why your expression has a value of 4.
Natasha_Volkova [10]
(2^3 +4) = 12

9-6=3

(2^3 +4) /(9-6)
12 /3
4
8 0
3 years ago
What is question 32 and can you please explain how to do it.
lukranit [14]

Answer is B.

5√3(5+2√6)=

25√3+10√18=

25√3+30√2

Note: I multiplied 10 by 3, because the prime factors of 18 are 3, 3 and 2 so I made a pair from the 3 and multipled and was left with 2.


5 0
3 years ago
Given b(x) = |x+41, what is b(-10)?<br> оооо.
AURORKA [14]

Answer:

31

Step-by-step explanation:

|-10 + 41| = 31

The absolute value of a number is ALWAYS positive.

I am joyous to assist you anytime.

7 0
3 years ago
• A researcher claims that less than 40% of U.S. cell phone owners use their phone for most of their online browsing. In a rando
antiseptic1488 [7]

Answer:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p > α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

Step-by-step explanation:

Set up hypotheses:

Null hypotheses = H₀: p = 0.40

Alternate hypotheses = H₁: p < 0.40

Determine the level of significance and Z-score:

Given level of significance = 1% = 0.01

Since it is a lower tailed test,

Z-score = -2.33 (lower tailed)

Determine type of test:

Since the alternate hypothesis states that less than 40% of U.S. cell phone owners use their phone for most of their online browsing, therefore we will use a lower tailed test.

Select the test statistic:  

Since the sample size is quite large (n > 30) therefore, we will use Z-distribution.

Set up decision rule:

Since it is a lower tailed test, using a Z statistic at a significance level of 1%

We Reject H₀ if Z < -1.645

We Reject H₀ if p ≤ α

Compute the test statistic:

$ Z =  \frac{\hat{p} - p}{ \sqrt{\frac{p(1-p)}{n} }}  $

$ Z =  \frac{0.31 - 0.40}{ \sqrt{\frac{0.40(1-0.40)}{100} }}  $

$ Z =  \frac{- 0.09}{ 0.048989 }  $

Z = - 1.84

From the z-table, the p-value corresponding to the test statistic -1.84 is

p = 0.03288

Conclusion:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p >  α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

8 0
3 years ago
Tara wants to lose 10 lbs. before spring break so that she looks and feels her best in a bathing suit. In the next 12 weeks, she
Harlamova29_29 [7]

is there a option for all of them?

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3 years ago
Read 2 more answers
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