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ASHA 777 [7]
3 years ago
7

Solve 15 < 4x on a number line

Mathematics
1 answer:
prisoha [69]3 years ago
7 0

Answer:

that shall answer 15 < 4x on a number line

Step-by-step explanation:

view image for more clarity

You might be interested in
Which of the following rational expressions has the domain restrictions X = -6 and x = 1?
Tcecarenko [31]

The domain of the function is possible values of independant varaible such that function is defined or have real values.

So the expression

\frac{(x+2)(x-3)}{(x-1)(x+6)}

is not defined for x = -6 and for x = 1, as expression becomes undefined for this values of x (Denominator becomes 0).

So answer is,

\frac{(x+2)(x-3)}{(x-1)(x+6)}

Option B is correct.

7 0
1 year ago
1.A five - layer of cake weighs half a kilogram per layer.In each layer,the cake is decorated with icing which weighs 15g.3 cand
julia-pushkina [17]

Answer:  2.815 kilograms

======================================================

Work Shown:

1 layer = 0.5 kg

5 layers = 5*(0.5 kg) = 2.5 kg

icing = 15 g = 15/1000 = 0.015 kg

1 candle = 100 g = 100/1000 = 0.1 kg

3 candles = 3*(0.1 kg) = 0.3 kg

------------------------

The five layers combine to 2.5 kg. On top of that we have 0.015 kg of icing, and then finally the three candles add 0.3 kg more weight.

The total weight is therefore: 2.5+0.015+0.3 = 2.815 kilograms

3 0
3 years ago
A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro
-Dominant- [34]

Answer:

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

Step-by-step explanation:

Data given and notation  

\bar X=1.6 represent the sample mean

s=0.46 represent the sample deviation

n=32 sample size  

\mu_o =1.35 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:  

Null hypothesis:\mu =1.35  

Alternative hypothesis:\mu \neq 1.35  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

P-value  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

5 0
3 years ago
#1 Write each fraction as a decimal Indicate
antiseptic1488 [7]
Decimal Exact form: 25/6
It is a terminating decimal
5 0
3 years ago
What is 1/4 times 2 -9​
krek1111 [17]

Answer:

-8.5

Step-by-step explanation:

1/4 times 2= 0.5

0.5-9=-8.5

Hope this helps :D

5 0
4 years ago
Read 2 more answers
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