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3 years ago

Solve the equation. Then check your solution. 1.6a=-9.12

Mathematics

2 answers:

balu736 [363]3 years ago

7 0

B.) a=-5.7
1.6(a)=-9.12
(a)=(-9.12)/(1.6)
a=-5.7

GaryK [48]3 years ago

5 0

Answer:

option (b) is correct.

a = - 5.7

Step-by-step explanation:

Given : equation 1.6a = - 9.12

We have to solve for a and choose the correct option from given options.

Consider the given equation 1.6 a = -9.12

Divide both side by 1.6

$\frac{1.6a}{1.6}=\frac{-9.12}{1.6}$

We have,

$a=\frac{-9.12}{1.6}$

Simplify, we have,

a = - 5.7

Thus, option (b) is correct.

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Find the slope of the line that contains these points (-2,5) (-4,11)

nevsk [136]

The first thing you have to do is:

subtract: 11-5

Then subtract: -4-(-2)

then you put it as a fraction: 6/-2 or -3

That -3 would be considered the slope.

So now you have: y=-3x+ b

So what you have to do is plug in one of the points into the equation to find out what b is.

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3 years ago

Suppose that in a random selection of 100 colored candies, 24% of them are blue. The candy company claims that the percentage

LuckyWell [14K]

Answer:

We conclude that the percentage of blue candies is equal to 23% at 0.01 significance level.

Step-by-step explanation:

We are given that in a random selection of 100 colored candies, 24% of them are blue.

The candy company claims that the percentage of blue candies is equal to 23%.

Let p = percentage of blue candies.

So, Null Hypothesis, $H_0$ : p = 23% {means that the percentage of blue candies is equal to 23%}

Alternate Hypothesis, $H_A$ : p $\neq$ 23% {means that the percentage of blue candies is different from 23%}

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

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

where, $\hat p$ = sample proportion of blue colored candies = 24%

n = sample of colored candies = 100

So, test statistics = $\frac{0.24-0.23}{\sqrt{\frac{0.24(1-0.24)}{100}} }$

= 0.234

The value of z test statistics is 0.234.

Now, at 0.01 significance level the z table gives critical values of -2.5758 and 2.5758 for two-tailed test.

Since our test statistics 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 percentage of blue candies is equal to 23%.

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3 years ago

How do I get (tan^2(x)-sin^2(x))/tan(x) equal to (sin^2(x))/cot(x)

shepuryov [24]

$LHS\\ \\ =\frac { \tan ^{ 2 }{ x-\sin ^{ 2 }{ x } } }{ \tan { x } } \\ \\ =\frac { 1 }{ \tan { x } } \left( \tan ^{ 2 }{ x-\sin ^{ 2 }{ x } } \right)$

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Assoli18 [71]

Answer:

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