Plz help; look at the picture

Solve the system by substitution.<br> y = -9x<br> y = 2x + 33

Gwar [14]

Answer:

$y = - 9x \\ theny = 2x + 33 \\ - 9x = 2x + 33 \\ - 9x - 2x = 33 \\ - 11x = 33 \\ x = \frac{ - 33}{11} = - 3 \\ x = - 3 \\then\\ y=-9×-3=21\\ thank \: you$

5 0

3 years ago

Select all the expressions whose value is larger than 100.

scZoUnD [109]

Answer:

120% of 100. 20% of 800. 500% of 400. 1% of 1,000

Hope this helps :)

4 0

2 years ago

The grocery store charges $10.35 for 5 bags of oranges.If someone bought 9 bags of oranges,how much would it cost then

Crank

It’s $18.63 , 10.35 divided by 5 = 2.07 , so 2.07 times 9 = $18.63

3 0

3 years ago

Questions 1-6 of 6 | Page 1 of 1

LenKa [72]

Answer:

Blank 1 = 53 + 90 = 143

Blank 2 = 143 - 229 = -86

Blank 3 = 0 - 86 = 86

7 0

3 years ago

The local swim team is considering offering a new semi-private class aimed at entry-level swimmers, but needs a minimum number o

SCORPION-xisa [38]

Answer:

The significance level is $\alpha=0.01$ and since we are conducting a right tailed test we need to find a critical value who accumulate 0.01 of the area in the right of the normal standard distribution and we got:

$z_{\alpha/2}= 2.326$

So we reject the null hypothesis is $z>2.326$

Step-by-step explanation:

For this case we define the random variable X as the number of entry-level swimmers and we are interested about the true population mean for this variable . On specific we want to test this:

Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

And the statistic is given by:

$z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

The significance level is $\alpha=0.01$ and since we are conducting a right tailed test we need to find a critical value who accumulate 0.01 of the area in the right of the normal standard distribution and we got:

$z_{\alpha/2}= 2.326$

So we reject the null hypothesis is $z>2.326$

7 0

3 years ago