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2 years ago

PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELPPPPPPPPPPPPPP

Mathematics

2 answers:

Arada [10]2 years ago

8 0

I think it is the 17Ab4

Margaret [11]2 years ago

4 0

Answer: I think the perimeter is 17ab4

Step-by-step explanation:

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Help someoneee!!!! ill give a brainliest

Serjik [45]

Answer:

-13.5

Step-by-step explanation:

-4/3w=18

You would need to get the w by itself on one side so the easiest way to do so would be to divide both sides my -4/3.

This would give you an answer of w= -27/2 or w= -13.5

8 0

2 years ago

Pls help me answer this :'( !!

alekssr [168]

Answer:

28 inches

Step-by-step explanation:

28×22/7×28

answer will come

7 0

2 years ago

Read 2 more answers

Let V=ℝ2 and let H be the subset of V of all points on the line 4x+3y=12. Is H a subspace of the vector space V?

sergejj [24]

Answer:

No, it isn't.

Step-by-step explanation:

We have $V=IR^{2}$ and let $H$ be the subset of $V$ of all points on the line

$4x+3y=12$

We need to find if $H$ is a subspace of the vector space $V$ .

In $IR^{2}$ all the possibilities for own subspace of the vector space $IR^{2}$ are :

$IR^{2}$ itself.

The vector $0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right]$

All lines in $IR^{2}$ that passes through the origin ( $0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right]$ )

We know that $H$ is the subset of $IR^{2}$ of all points on the line $4x+3y=12$

If we look at the equation, the point $\left[\begin{array}{c}0&0\end{array}\right]$ doesn't verify it because :

$4x+3y=12\\4(0)+3(0)=12\\0=12$

Which is an absurd. Therefore, $H$ doesn't contain the origin (and $H$ is a line in $IR^{2}$ ). Finally, it can't be a vector space of $V=IR^{2}$

8 0

2 years ago

What is 500000 and 60000000 as a decimal

Juliette [100K]

The only way to make a decimal out of either of those numbers is to write a decimal point with some zeros after it.

500,000 = 500,000.000

60,000,000 = 60,000,000.00000

etc.

3 0

3 years ago

Suppose a consumer group suspects that the proportion of households that have three cell phones NOT known to be 30%. A cell phon

leva [86]

Answer:

We conclude that the actual percentage of households is equal to 30%.

Step-by-step explanation:

We are given that a consumer group suspects that the proportion of households that have three cell phones NOT known to be 30%.

Their marketing people survey 150 households with the result that 43 of the households have three cell phones.

Let p = proportion of households that have three cell phones NOT known.

So, Null Hypothesis, $H_0$ : p = 30% {means that the actual percentage of households is equal to 30%}

Alternate Hypothesis, $H_A$ : p $\neq$ 30% {means that the actual percentage of households different from 30%}

The test statistics that would be used here One-sample z-test for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of households having three cell phones = $\frac{43}{150}$ = 0.29

n = sample of households = 150

So, the test statistics = $\frac{0.29-0.30}{\sqrt{\frac{0.30(1-0.30)}{150} } }$

= -0.27

The value of z test statistic is -0.27.

Also, P-value of the test statistics is given by;

P-value = P(Z < -0.27) = 1 - P(Z $\leq$ 0.27)

= 1 - 0.6064 = 0.3936

Now, at 1% significance level the z table gives critical value of -2.58 and 2.58 for two-tailed test.

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the actual percentage of households is equal to 30%.

4 0

3 years ago

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